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Sustainability in Math Activities
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Population Growth, Ecological Footprints, and Overshoot part of Activities
Rikki Wagstrom
In this activity, students develop and apply linear, exponential, and rational functions to explore past and projected U.S. population growth, carbon footprint trend, ecological overshoot, and
effectiveness of hypothetical carbon dioxide reduction initiatives.
A Monarchy Deposed: The Demise of the Monarch Butterfly part of Activities
Daniel Abel
Monarch butterflies (scientific name: Danaus plexippus) migrate annually to forests in central Mexico from Canada and California. Those surviving the 1200 - 2800 mile migration overwinter in Mexico.
In this activity, students will learn about the conservation biology of monarch butterflies, threats to their survival, the implications of their potential extinction, and ways to protect the
Plastic Waste Production part of Activities
Karen Bliss
In this exercise, students will use data to predict the amount of plastic waste in the next ten years.
How much energy do you save by doubling insulation? part of Activities
Joseph Skufca
Students will be provided the governing equation for steady state heat transfer across a surface. They will use that equation to explore the effect of changing the insulation value on the amount of
energy used.
Arctic Sea Ice Extent part of Activities
Bill Bauldry
Student teams investigate Arctic Sea Ice by analyzing actual data and making predictions. A worthwhile extension is to predict the first year that the Arctic Ocean will be ice free.
How Big is Your Breakfast Footprint? part of Activities
Ben Galluzzo, Shippensburg University
Calculation of a carbon footprint resulting from common breakfast choices illustrates the importance of contextualization.
The Costs of Your Commute: Your Money, Your Time, and the Earth part of Activities
Charlie Buehrle
This activity has students investigate their own cost, CO2 output, and time for commuting. They then compare their commute to an environmentally conscious alternative by using comparable metrics.
Should I Unplug? part of Activities
Lori Carmack
Short Description goes here: Short DescriptionThis section should be a distillation of the summary above. This description will be displayed in search returns. The optimal length for this description
is on the order of 1-2 sentences. Short Description goes here: | {"url":"https://serc.carleton.edu/sisl/2012workshop/activities.html?q1=sercvocabs__214%3A4","timestamp":"2014-04-23T12:32:58Z","content_type":null,"content_length":"23364","record_id":"<urn:uuid:bbb6ea99-993c-4d42-8076-dead1f90607d>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00216-ip-10-147-4-33.ec2.internal.warc.gz"} |
Test for Series Convergence/Divergence
April 29th 2010, 04:32 PM #1
Dec 2008
Test for Series Convergence/Divergence
Hi all, I have a series sum from n = 1 to infinity for n!/n^n.
I've used the ratio test only to come up with a 1 which is obviously inconclusive. I tried the root test and also came up with a 1 though it's quite possible I mixed this up. I feel as though it
has to do with either direct comparison or limit comparison but I'm drawing blanks and any help would be greatly appreciated.
Hi all, I have a series sum from n = 1 to infinity for n!/n^n.
I've used the ratio test only to come up with a 1 which is obviously inconclusive. I tried the root test and also came up with a 1 though it's quite possible I mixed this up. I feel as though it
has to do with either direct comparison or limit comparison but I'm drawing blanks and any help would be greatly appreciated.
let me guess ...
you did the ratio test and came up with
$\lim_{n \to \infty} \left(\frac{n}{n+1}\right)^n$
and you think the limit is 1 , right?
I didn't even recognize it as 1/e with it written as (n^n)/(n+1)^n
Thanks a ton.
Edit 2: You beat me to it, surprise = 0
Last edited by zo1971so; April 29th 2010 at 05:19 PM. Reason: oops 1/e not e
For those of us who are a little slow (i.e., me), I've written out a detailed proof
We can use the ratio test to test for this series' convergence.
$L=\lim_{n\to\infty}\frac{\frac{\left(n+1\right)!}{ \left(n+1\right)^{n+1}}}{\frac{n!}{n^n}}$
$=\lim_{n\to\infty}\frac{\left(n+1\right)!}{\left(n +1\right)^{n+1}}\frac{n^n}{n!}$
$=\lim_{n\to\infty}\frac{\frac{n+1}{n^n}}{\left(n+1 \right)^{n+1}}$
The ratio test specifies that if $L<1$ the series converges absolutely. Thus, this series converges absolutely.
Last edited by lovek323; April 29th 2010 at 06:08 PM. Reason: Added link
April 29th 2010, 04:53 PM #2
April 29th 2010, 05:07 PM #3
Dec 2008
April 29th 2010, 05:14 PM #4
April 29th 2010, 06:08 PM #5
Apr 2010
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Finance[ForwardCurve] - create new zero curve based on instantaneous forward rates
Calling Sequence
ForwardCurve(rate, opts)
ForwardCurve(times, rates, opts)
ForwardCurve(dates, rates, opts)
rate - real constant, algebraic expression, or a procedure; forward rate
times - list or Vector; times (in years)
rates - list or Vector; forward rates
dates - list; dates
opts - equations of the form option = value where option is one of compounding, daycounter, interpolation, or referencedate; specify options for the ForwardCurve command
• compounding = Simple, Continuous, Annual, Semiannual, EveryFourthMonth, Quarterly, Bimonthly, Monthly, SimpleThenAnnual, SimpleThenSemiannual, SimpleThenEveryFourthMonth, SimpleThenQuarterly,
SimpleThenBimonthly, or SimpleThenMonthly -- This option specifies the compounding type for the given interest rate(s).
• daycounter = Actual360, Actual365Fixed, AFB, Bond, Euro, Historical, ISDA, ISMA, OneDay, Simple, Thirty360BondBasis, Thirty360EuroBondBasis, Thirty360European, Thirty360Italian, Thirty360USA, or a
day counter data structure -- This option specifies the convention used to convert the amount of time between two dates to year fractions.
• interpolation = BackwardFlat, Cubic, ForwardFlat, Linear, or LogLinear -- This option specifies the type of interpolation used to build a forward curve from a discrete set of rates. The LogLinear
interpolation is used by default.
• referencedate = date in any of the formats recognized by the ParseDate command -- This option specifies the reference date (when the discount factor is equal to 1).
• The ForwardCurve command creates a new yield curve based on the specified instantaneous forward rates; the resulting curve is represented as a module. This module can be passed to other commands
of the Finance package that expect a yield term structure as one of the parameters; it can also be used as if it were a procedure. Assume for example that the module returned by ForwardCurve was
assigned to the name R. Then for any positive constant t, will return a zero rate for the maturity t based on the term structure R. If d is a date given in any of the formats recognized by the
ParseDate command, then the command will return the forward rate for the corresponding maturity.
• The ForwardCurve(rate, opts) command creates a zero curve based on the specified interest rate. The parameter rate can be either a real constant, a Maple procedure, or an algebraic expression. If
rate is a real constant, then the ForwardCurve command contracts a flat term structure based on the specified interest rate. If rate is a procedure, it should accept one parameter (the time) and
return the corresponding rate as a floating-point number. Finally, if rate is an algebraic expression, it should depend on a single variable. This variable will be taken as time.
• The ForwardCurve(times, rates, opts) and ForwardCurve(dates, rates, opts) commands create a term structure based on piecewise interpolation of specified forward rates. The parameters rates and
times can be either a list or a Vector containing numeric values and must have the same number of elements. The parameter dates is a list of dates in one of the formats recognized by ParseDate.
• By default all rates are assumed to be based on continuous compounding. The compounding option can be used to specify the desired compounding type for the given rates. If the value of the
compounding option is different from Continuous, all the given rates will be converted to the corresponding continuous rates.
• Objects created using the ForwardCurve command will be of Maple type YieldTermStructure.
• The Finance[ForwardCurve] command was introduced in Maple 15.
• For more information on Maple 15 changes, see Updates in Maple 15.
In this example, create a flat forward curve with reference date set to January 5, 2005.
In this example, create a forward curve with the same parameters as above but assume that the interest rate is based on the monthly compounding.
Note that in the previous example only the rates passed to the ZeroCurve constructor are assumed to be monthly compounded. Whenever we evaluate forwardcurve2, the returned rate is always continuous.
In this example, create a forward curve based on a piecewise interpolation of forward rates. Use the default interpolation.
Construct a forward yield term structure based on a Maple function.
See Also
Finance[BlackScholesProcess], Finance[DiscountCurve], Finance[DiscountFactor], Finance[ForwardRate], Finance[MertonJumpDiffusion], Finance[ParRate], Finance[ZeroCurve], Finance[ZeroRate]
Brigo, D., Mercurio, F., Interest Rate Models: Theory and Practice. New York: Springer-Verlag, 2001.
Hull, J., Options, Futures, and Other Derivatives, 5th. edition. Upper Saddle River, New Jersey: Prentice Hall, 2003.
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Compact sets in TVS
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
Let $K$ be a compact subset of a Hausdorff topological vector space. Is it true that $\bigcap_{n\in \mathbb{N}}\frac{1}{n}K$ is either empty of or is a set consisting of
the origin only?
up vote 1 down vote
favorite fa.functional-analysis topological-vector-spaces
add comment
Let $K$ be a compact subset of a Hausdorff topological vector space. Is it true that $\bigcap_{n\in \mathbb{N}}\frac{1}{n}K$ is either empty of or is a set consisting of the origin only?
I'll assume this is a TVS over $\mathbb{R}$ or $\mathbb{C}$, and not over some other local field.
up vote 6 If $x$ is nonzero and belongs to $\frac1{n}K$ for every $n \geq 1$, then $nx \in K$ for each $n$. If $L$ is the line through $x$, then the subspace topology coincides with the standard
down vote topology on the ground field by the TVS axioms, and $L \cap K$ is a compact subset of the line which is unbounded since it contains all the $nx$. This gives a contradiction.
add comment
I'll assume this is a TVS over $\mathbb{R}$ or $\mathbb{C}$, and not over some other local field.
If $x$ is nonzero and belongs to $\frac1{n}K$ for every $n \geq 1$, then $nx \in K$ for each $n$. If $L$ is the line through $x$, then the subspace topology coincides with the standard topology on
the ground field by the TVS axioms, and $L \cap K$ is a compact subset of the line which is unbounded since it contains all the $nx$. This gives a contradiction. | {"url":"http://mathoverflow.net/questions/67589/compact-sets-in-tvs/67591","timestamp":"2014-04-20T01:41:55Z","content_type":null,"content_length":"54564","record_id":"<urn:uuid:f7d71ade-01e2-4eeb-95ec-3ac56736602e>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00463-ip-10-147-4-33.ec2.internal.warc.gz"} |
Rest Haven, GA Algebra Tutor
Find a Rest Haven, GA Algebra Tutor
I have a BS and MS in Physics from Georgia Tech and a Ph.D. in Mathematics from Carnegie Mellon University. I worked for 30+ years as an applied mathematician for Westinghouse in Pittsburgh.
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I need help in this Probabilistic Model
October 16th 2009, 02:58 PM
I need help with this Probabilistic Model
Alice and Bob each choose at random a number between zero and two. We assume a uniform probability law under which the probability of an event is proportional to its area. Consider the following
A: The magnitude of the difference of the two numbers is greater than 0.44.
B: At least one of the numbers is greater than 0.44.
C: The two numbers are equal.
D: Alice's number is greater than 0.44.
Find the following probabilities.
P(A) =
P(B) =
P(A∩B) =
P(C) =
P(D) =
P(A∩D) =
October 18th 2009, 04:20 AM
does this work?
2 >|x-y| >0.44 = 1.56/2 = 0.78 = P(A) ?
For (B) = use law of total probability = (P(over)P(under))+(P(under)P(over))
=(1.56/2)x(0.44/2)+(0.44/2)+(1.56/2) = .34 | {"url":"http://mathhelpforum.com/advanced-statistics/108468-i-need-help-probabilistic-model-print.html","timestamp":"2014-04-21T14:43:07Z","content_type":null,"content_length":"4087","record_id":"<urn:uuid:a442e19b-678a-4007-ab5f-e577995e7002>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00509-ip-10-147-4-33.ec2.internal.warc.gz"} |
Negligible function question.
There is a question:
In constructing obfuscators for point-functions theory, there is a statement that there exists a polynomial-time computable permutation
[tex]pi : B^n -> B^n[/tex] and a constant c such that for every polynomial s(n) and every adversary A of size s for all sufficiently large n,
[tex] Prob[A(pi(x)) = x] <= S(n)^c/2^n [/tex]
I am trying to prove that
where s is a polynomial and c is a constant, is also a negligible function.
Could anyone help me with that? | {"url":"http://www.physicsforums.com/showthread.php?p=937799","timestamp":"2014-04-21T02:12:44Z","content_type":null,"content_length":"19844","record_id":"<urn:uuid:bd6c661d-8db0-47a6-a494-8c14b0c41a68>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00546-ip-10-147-4-33.ec2.internal.warc.gz"} |
Newton Center Algebra Tutor
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Image SXM
Image analysis software NIH Image extended to handle scanning microscope images
Date added:30 Jan 2012
Tags: image analysis, image processing, NIH Image, scanning microscopy, AFM, LSM, SAM, SEM, SFM, SLM, SNOM, SPM, STM,
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Advanced Graphing Calculator 3D Mac
Plot high quality 2D and 3D graphs of math equations and coordinates tables.
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Graphing Calculator 3D Mac
Plot high quality 2D and 3D graphs of math equations and coordinates tables.
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Statistical analysis on a Mac made easy with StatPlus:mac and Microsoft Excel
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it has been a while since ive done these so i just need a refresher. im looking for the heat of hydration of lithium chloride and sodium chloride. compound Lattice Energy deltaH soln LiCl -834 kj/mol
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MathGroup Archive: December 2004 [00551]
[Date Index] [Thread Index] [Author Index]
Re: How input stacked characters with vertical bar
• To: mathgroup at smc.vnet.net
• Subject: [mg53061] Re: How input stacked characters with vertical bar
• From: "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk>
• Date: Wed, 22 Dec 2004 04:52:53 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
Here is a Text cell expression that does what you want.
FractionBox["g", "h"]], "\[RightBracketingBar]"}], "a",
"b"], TraditionalForm]]]], "Text"]
The trick is to use the left and right bracketing bars (these are on the
BasicTypsetting palette just below the corresponding left and right angle
brackets). You then need to colour the left hand bracketing bar white (so it
disappears). It is not OK to simply delete it, because the right hand
bracketing bar then appears purple (because it is lonely without its
partner!), and this colouration survives all the way to the final printed
Steve Luttrell
"Murray Eisenberg" <murray at math.umass.edu> wrote in message
news:cq8tl2$h1h$1 at smc.vnet.net...
> OK, that does stack th subscript and superscript -- but they are not
> placed where proper mathematical notation says they should appear with
> such a vertical bar.
> I've tried both the | key on the keyboard as well as the
> VerticalSeparator symbol (on the Complete Characters palette). And
> those symbols do (almost) automatically expand to the height of what's
> immediately to their left, e.g., a stacked:
> 2
> x
> ---
> 2
> But the subscript and superscript I type then are NOT located near the
> top and bottom of the vertical bar, where they ought to appear. Rather,
> they are quite close together vertically, as if they were the subscript
> and superscript on a single letter.
> Paul Abbott wrote:
>> In article <cpekge$74l$1 at smc.vnet.net>,
>> Murray Eisenberg <murray at math.umass.edu> wrote:
>>>In a text cell, as part of an in-line mathematical expression, how do I
>>>type a stacked pair of expressions, both the same size. For example,
>>>the stacked b above a, next to a long vertical bar, as you would see in
>>>a statement of the Fundamental Theorem of Calculus that expresses the
>>>definite integral as F(b) - F(a).
>> Just as you do for an integral or a sum. Use
>> \[ControlKey] -
>> (subscript) for the lower limit and toggle to the upper limit using
>> \[ControlKey] 5
>> These are in the
>> Edit | Expression Input
>> menu.
>>>And which is the appropriate symbol for that vertical bar?
>> \[VerticalSeparator]
>> The keyboard shortcut is
>> esc | esc
>>>Is there some way to force it to expand vertically to the extent of the
>>>stacked expressions next to it, or must one do so manually.
>> Select the VerticalSeparator, open the Option Inspector, type in span,
>> hit return, and set the SpanMaxSize to Infinity.
>> Here is an example of such a cell:
>> Cell[BoxData[FormBox[RowBox[{RowBox[{"F", "(", "x", ")"}],
>> SubsuperscriptBox[StyleBox["\[VerticalSeparator]",
>> SpanMaxSize->Infinity], "a", "b"]}],
>> TraditionalForm]], "Input"
>> ]
>> Cheers,
>> Paul
> --
> Murray Eisenberg murray at math.umass.edu
> Mathematics & Statistics Dept.
> Lederle Graduate Research Tower phone 413 549-1020 (H)
> University of Massachusetts 413 545-2859 (W)
> 710 North Pleasant Street fax 413 545-1801
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[Numpy-discussion] Raveling, reshape order keyword unnecessarily confuses index and memory ordering
josef.pktd@gmai... josef.pktd@gmai...
Tue Apr 2 13:37:39 CDT 2013
On Tue, Apr 2, 2013 at 2:04 PM, Matthew Brett <matthew.brett@gmail.com> wrote:
> Hi,
> On Tue, Apr 2, 2013 at 12:29 PM, Chris Barker - NOAA Federal
> <chris.barker@noaa.gov> wrote:
>> On Mon, Apr 1, 2013 at 10:15 PM, Matthew Brett <matthew.brett@gmail.com> wrote:
>>> Thank you for the compliment, it's more enjoyable than other potential
>>> explanations of my confusion (sigh).
>>> But, I don't think that is the explanation.
>> well, the core explanation is these are difficult and intertwined
>> concepts...And yes, better names and better docs can help.
>>> Last, as soon as we came to the distinction between index order and
>>> memory layout, it was clear.
>>> We all agreed that this was an important distinction that would
>>> improve numpy if we made it.
>> yup.
>>> I think you agree that there is potential for confusion, and there
>>> doesn't seem any reason to continue with that confusion if we can come
>>> up with a clearer name.
>> well, changing an API is not to be taken lightly -- we are not
>> discussion how we'd do it if we were to start from fresh here. So any
>> change should make things enough better that it is worth dealing with
>> the process of teh change.
> Yes, for sure. I was only trying to point out that we are not talking
> about breaking backwards compatibility.
>>> So here is a compromise proposal.
>>> * Preferring the names 'c-style' and 'f-style' for the indexing order
>>> case (ravel, reshape, flatiter)
>>> * Leaving 'C" and 'F' as functional shortcuts, so there is no possible
>>> backwards-compatibility problem.
>> seems reasonable enough -- though even with the backward
>> compatibility, users will be faces with many, many older examples and
>> docs that use "C' and 'F', while the new ones refer to the new names
>> -- might this be cause for even more confusion (at least for a few
>> years...)
> I doubt it would be 'even more' confusion. They would only have to
> read the docstrings to work out what is meant, and I believe, with
> better names, they'd be less likely to fall into the traps I fell
> into, at least.
>> leaving me with an equivocal +0 on that ....
>> antoher thought:
>> """
>> Definition: np.ravel(a, order='C')
>> A 1-D array, containing the elements of the input, is returned. A copy is
>> made only if needed.
>> Parameters
>> ----------
>> a : array_like
>> Input array. The elements in ``a`` are read in the order specified by
>> `order`, and packed as a 1-D array.
>> order : {'C','F', 'A', 'K'}, optional
>> The elements of ``a`` are read in this order. 'C' means to view
>> the elements in C (row-major) order. 'F' means to view the elements
>> in Fortran (column-major) order. 'A' means to view the elements
>> in 'F' order if a is Fortran contiguous, 'C' order otherwise.
>> 'K' means to view the elements in the order they occur in memory,
>> except for reversing the data when strides are negative.
>> By default, 'C' order is used.
>> """
>> Does ravel need to support the 'A' and 'K' options? It's kind of an
>> advanced use, and really more suited to .view(), perhaps?
>> What I'm getting at is that this version of ravel() conflates the two
>> concepts: virtual ordering and memory ordering in one function --
>> maybe they should be considered as two different functions altogether
>> -- I think that would make for less confusion.
> I think it would conceal the confusion only. If we don't have 'A'
> and 'K' in there, it allows us to keep the dream of a world where 'C"
> only refers to index ordering, but *only for this docstring*. As
> soon as somebody does ``np.array(arr, order='C')`` they will find
> themselves in conceptual trouble again.
I still don't see why order is not a general concept, whether it
refers to memory or indexing/iterating.
The qualifier can be made clear in the docstrings (or from the context).
It's all over the documentation:
we can iterate in F-order over an array that is in C-order (*), or vice-versa
(*) or just some strides
pure shape
shape and copy
> Cheers,
> Matthew
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion@scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
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Biggest Test of Benford's Law?
Earlier this week a swapped a couple of emails with someone at Google involved with looking at web page statistics on the Google database. They probably have the one of the largestest corpus of
textual data in the world and it's a mixture of data from a wide variety of sources. This makes it an ideal candidate for testing
Benford's law
and that's what I'm suggesting they do. The published tests I've seen so far are pretty small but it'd be fun to see the results from, say, a billion web pages. Coincidentally
Boing Boing
just had a linky to a New York Times
on it form 1998. Anyway...I asked the guy to email me if they carry out the test. It's probably pretty low down on their list of priorities. So of course I'll report back.
22 comments:
Mike said...
Hopefully you read about my smaller-scale test of Benford's Law with Google. Check it out here:
sigfpe said...
Nice! But I'd love to see where we get wih a million numbers!
sigfpe said...
Did I say a million? I mean a billion!
Shay said...
I'm not sure how anyone can say "nobody knows why" numbers starting with 1 are more common. I'm not a mathematician but it seems quite obvious to me.
As an example, there are only a few cities in the world with more than 10 million people, a handful with 5 million, and maybe 100 or 200 with 1 million. There are countless thousands of small
communities with 1,000 people. As the numbers get bigger, the number of places with that number gets smaller.
So if we are talking six digits, more places will have 100,000-200,000 people than 200,000-300,000 people, and so on. This would be true of most things that require a number - smaller numbers are
more plentiful.
Moshe said...
Benford's Law does not seem to me to be all that complex to understand.
Assuming most numbers in this world are NOT random, that is to say, individually, they are part of a larger sequence, then 1 would be the most used number followed by 2 etc.
Taking addresses as an example, the address 239 XXX street is necessarily preceeded by 238, 237 and so on. Crucially, the addresses all START at 1 as do most ordered things (imagine giving the
first person in a lineup the number 78364)and thus, there are more 1's than any other number used in our daily lives.
Benford's law solved. I now turn my attention to figuring out how ANYONE can like black licorice.
sigfpe said...
But (1) you aren't explaining the precise logarithmic distribution and (2) this distribution applies to things that aren't sequences such as prices in a supermarket or the lengths of files on a
Chris said...
yes, I think it does explain the logarithmic distribution.
Most numbers out there are sequences, starting from 1 (why would you use random numbers?) There may be examples where numbers are more random (computer file sizes) but these will be heavily
outweighed by the number of sequences out there.
Also, computer file sizes could be seen as something that increases sequentially. You write code, the file size gets larger. You add more code it gets larger still. You have to go through file
size 1000 before getting to 2000 before getting to 3000. You stop at some point when the file is complete. Overall more will have stopped at 1000+ than at 2000+ than at 3000+ etc (because you
have to go through these stages before reaching the larger sizes. Therefore there will be the logarithmic distribution exactly as shown.
Supermarket prices are the same. You have far more low priced items than high priced items. To become a high priced item you have to go through all the low prices first, therefore a prices is
more likely to start with a 1 than a 2 than a 3 etc
sigfpe said...
The fact that you have to go through files of size 1000 to get to a file of size 2000 doesn't entail that there are more files of size 1000. For example, if I wrote a program to spew out millions
of files of length 2000 I'd expect to see no files of length 1000 in my filesystem.
Supermarkets aren't forced to give products a low price which is then increased. New products come on the market all the time. Benford's law will apply to the set of all new products that have
just come onto the market. And Benford's law would hold in market prices even if we were going through a period of deflation. Benford's law will hold for 1/price when there is inflation, ie. now.
None of what you say explains why as you grow the dataset the distribution gets closer and closer to precisely logarithmic as opposed to any other distribution that decreases with digit size.
Chris said...
'For example, if I wrote a program to spew out millions of files of length 2000 I'd expect to see no files of length 1000 in my filesystem.'
not really the best example is it? You are setting out to artificially make an exception that disproves the rule.
Yes, there may be cases where particular files are in a particular size range (for instance I deal with huge numbers of low res jpg images which will tend to a particular size of maybe 50000 -
80000k in size) but these are exceptions. The vast majority of files will grow in size with the amount of data they contain, and will vary enormously in size, but the large files will have had to
have been small files before they became big files.
open a random folder on your computer and check the file sizes...
also, whenever a list of file sizes increases to the next 'power' (if that is the right term) it will automatically be tipped dramatically back towards confirming Benfords law
foe instance, if the file sizes increase steadily, as soon as they go from 9999k to 10000k, 10 times as many files from then on will start with a 1 than started with any earlier number. It will
start to even out as the number increases, but the same thing will happen when it hits 100000k etc
as lists of numbers can stop at any point within this progression, overall the balance will be towards the lower numbers (as you alwasy have to have the lower numbers before the higher numbers)
I suspect supermarket prices are probably not a good example anyway, because prices will often be set at 99.99 or 99.95 rather than tipping it over to a psychologically higher number.
However, supermarket prices, and other exceptions, will be overwhelmingly drowned out by most number sequences which do not have these artificial effects on their progression.
sigfpe said...
> You are setting out to artificially
> make an exception that disproves the
> rule.
Exactly. That's the whole point of a counterexample - to contrive a case that demonstrates that a rule doesn't hold.
Sequences don't imply a logarithmic distribution. If I toss a coin 1,000 times and count how many times I get heads I have a sequence. I'll start with one head, go to two, then three and so on.
The final distribution is approximately Gaussian centred on 500, nothing like a logarithmic distribution.
Conversely, logarithmic distributions don't imply sequences. For example, the reciprocals of prices (ie. 1/price) in a supermarket also have a logarithmic distribution and get you can't argue
that recirprocal prices form a sequence.
So there is no necessary connection between a logarithmic distribution and having a sequence.
If you search Benford's Law on, of all places, Google, you'll get an explanation of why the distribution is exactly the way it is. When I searched it was the first link, but such things change.
The basic principles discussed here are correct but they don't explain why in a large enough distribution there are about 30% 1's instead of, say, 25% 1's. The website I found (http://
mathworld.wolfram.com/BenfordsLaw.html) uses calculus to develope a formula:
where D is the first digit.
Now I'm not saying that I understand how the proof works, as I'm a little rusty on such things and it wasn't very detailed, but I almost understand it, so any claims that nobody knows why the law
works stike me as quite doubtful.
Also, it doesn't work with random numbers or small sets heavily influenced by psychology, it is a distribution that works better the larger and more diverse the sample, particularly with numbers
that arise naturally. Counterexamples are irrelevant unless they fall on a scale where this is supposed to work.
sigfpe said...
> but I almost understand it, so any claims that nobody knows why the law works stike me as quite doubtful.
The problem is that the argument on the Mathworld web site isn't a valid proof for two reasons that the article itself points out: (1) P(x)=1/x cannot possibly be a probability distribution as
its integral diverges (ie. the sum of probabilities is infinite, not one) and (2) Benford's law applies to mixtures of data from a variety of sources. The resolution of these issues only came in
> Counterexamples are irrelevant unless they fall on a scale where this is supposed to work.
I don't know what you mean by "scale" here. I could generate a trillion files, each a billion bytes long and there'd be no files less than a billion bytes long. So I presume you're not talking
about either the size of the numbers or the number of numbers.
T-Bon3 said...
The most important reason is that numbers are commonly used as measurements of quantities.
Think about this example, try it in excel if you like:
Take an amount of money as a number, say $1,000. Repeatedly increase it by 5% until you get to $10,000 now count the number of numbers that start with each digit.
Yo will get 15 1's, 8 2's, 6 3's, 4 4's, 4 5's, 3 6's, 3 7's, 3 8's, and 2 9's
In the range of amounts $1,000-$9,999 the numbers in the $1000-$2000 range are more prevalent because this is actually a bigger proportion in terms of the % changes. As a lot of economic numbers
(the article is about the IRS, and tax returns, not phone numbers or other lists) change in a 'percentage' way not a fixed increase (price rises, wages, stocks, production, etc.) this
distribution is inevitable.
sigfpe said...
You describe a process that gives you a logarithmic distribution (the correct one!) for a similar reason to Benford's law, but it isn't *exactly* the same. For example, Benford's law will hold if
you look at something like the densities of materials in a chemical data book. These aren't dynamic values that increment repeatedly by a percentage.
So random, yet so interesting. I am not a mathematician either, but I did stay at a Holiday Inn Express last night.
It seems to me that ultimately this could be explained by adding in the human element here -- statistics of human behavior and perception are very often observed to fit neatly into one or more
power laws, many of which are logarithmic.
See Wikipedia, where "[power laws] appear to fit such disparate phenomena as the popularity of websites, the wealth of individuals, the popularity of given names, and the frequency of words in
"Author Philip Ball has argued that the same power law relationships that are evident in phase transitions also apply to various manifestations of collective human behaviour."
If I remember my statistical analysis of human behavior class correctly, perception of stimuli, including sound waves, is most often expressed logarithmicly.
For instance, the perceived difference between a sound wave of 10db and a sound wave of 100db is actually the same as the difference between waves of 100,000db and 1,000,000db.
Because of this, the volume dials on most stereos are actually calibrated acording to a logarithmic scale, e.g. an arbitrary volume setting of "2" on your speakers has an actual amplitude
difference of 10x the volume of "1".
Here, we have simply adjusted the means by which we are measuring the sound -- the volume dial -- to best fit our perception.
In this way, we might assume that other systems by which we measure and record data also reflect this, including the decimal system of numbers, which is incidentally a base-10 system where
position of a numerical symbol expreses the value of that symbol in terms of the exponential values of the base.
I am not suggesting that a table of molecular weights or sizes of computer files are somehow 'biased' by our perception, just that the systems of measurement and numbers used to express data are
really just human constructs adopted acording to their utility.
This utility would in theory include the representation of the world according to logarithmic perception. I could be totally off base, but it seems to me that, like the volume dial, the decimal
system is just another way that we as perceiving entities have devised to express the world around us.
sigfpe said...
There is some truth to what you say - partly for the obvious reason that we are talking about first digits, which is obviously an artifact of our number system. A culture that used a logarithmic
system, on the other hand, might find that the equivalent of their first digit was uniformly distributed.
BTW One of my pet irks is that many mp3 playing applications don't mimic the logarithmic volume controls on real stereos meaning that the useful range of volume control is all compressed at one
PS How come everyone suddenly stumbling on this year old blog entry of mine?
"How come everyone suddenly stumbling on this year old blog entry of mine?"
It may be because of this link being passed around: http://www.thecleverest.com/countdown.swf
saber tooth owl said...
Benford's Law is fascinating, and makes more sense as you examine it further.
For denotations of time, the leading numberal one is highly favored, since four of the 12 possibilities start with that number (1, 10, 11 & 12)
On military times, however, it's skewed toward zero, obviously since ten out of 24 will start there. In actuality, most usage of military times will probably front-load with normal duty hours,
say between 0700 to 1700, fiddling it back a bit, since logbooks and such will have more entries during those hours, and more activities take place during those hours.
MetaJon said...
plug this into a binary to text translator for the solution.
sigfpe said...
"Benford's Law"
Smaller amounts are more common in our world. Why, you ask? Larger objects/amounts are harder to maintain and harder to come by, because of the amount of space required to contain it and the
amount of energy required to handle it.
As I've seen mentioned elsewhere, one facet of this is the way units are defined.
If there's a typical amount of something, it makes sense to create a unit size near that amount.
If not, another unit is often chosen. And with the metric system, it's the natural inclination to change the unit (even to a non-base unit) and drop the precision.
928m will be described as 1km
I think that humans influence this far more than some of the people discussing this realize. | {"url":"http://blog.sigfpe.com/2006/01/biggest-test-of-benfords-law.html?showComment=1170098940000","timestamp":"2014-04-18T20:45:05Z","content_type":null,"content_length":"93970","record_id":"<urn:uuid:d4508c96-5f96-48fd-92c3-f802066fe5a6>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00499-ip-10-147-4-33.ec2.internal.warc.gz"} |
1. CONTEXT
Were we to choose at random a hundred members of the IAU and ask each of them to tell us the value of the Hubble constant and how it is measured, few if any would refer to gravitational lens time
delay measurements. Most would instead refer to observations of Cepheid variables with HST or to observations of the CMB power spectrum with WMAP. These set the de facto standard against which time
delay estimates must be evaluated.
The Cepheids give a Hubble constant of 72 km/s/Mpc with a 10% uncertainty [Freedman et al. (2001)], a result that most of our randomly chosen astronomers would find relatively straightforward. But
only a handful of them would be able to tell us how the CMB power spectrum yields a measurement of the Hubble constant.
In his opening talk, David SPERGEL told us that three numbers are determined to high accuracy by the CMB power spectrum: the height of the first peak, the contrast between the heights of the even and
odd peaks, and the distance between peaks. Ask anyone who calls himself a cosmologist how many free parameters his world model has and the number N will be larger than three. Spergel's three numbers
constrain combinations of those N parameters, but do not, in particular, tightly constrain the Hubble constant. One must supplement the CMB either with observations, or alternatively, with
non-observational constraints.
The effects of supplementing the observed CMB power spectrum with other observations, and of adopting (or declining to adopt) non-observational constraints, are shown in table 1. All of the numbers
shown are taken from the analysis by [Tegmark et al. (2004b)]. They use the first year data from the Wilkinson Microwave Anisotropy Probe [Bennett et al. (2003)], the second data release of the Sloan
Digital Sky Survey [Abazajian et al. (2004)], the [Tonry et al. (2003)], data for high redshift Type Ia supernovae, and data from 6 other CMB experiments: Boomerang, DASI, MAXIMA, VSA, CBI and ACBAR.
Table 1. The CMB power spectrum and the Hubble constant
very-nearly flat perfectly flat
observations [tot] 100h observations [tot] 100h
WMAP1 1.086^+0.057[-0.128] 50^+16[-13] WMAP1 1 74^+18[-7]
add SDSS2 1.058^+0.039[-0.041] 55^+9[-6] add SDSS2 1 70^+4[-3]
add SNae 1.054^+0.048[-0.041] 60^+9[-6] add other CMB 1 69^+3[-3]
We see from the first line of table 1 that the value of the Hubble constant derived from the WMAP1 data alone is a factor of 1.5 or 2.5 more uncertain than the Cepheid result, depending upon whether
the universe is taken to be perfectly flat or only very-nearly flat. Combining the galaxy power spectrum as measured with SDSS2 [Tegmark et al. (2004a)] reduces the uncertainties by factors of 3 and
2, respectively. In the very-nearly flat case, adding type Ia supernovae does little to change the uncertainty but shifts the value of H[0] closer to the Cepheid value. In the perfectly flat case,
adding other CMB measurements reduces the uncertainty in H[0] by another factor of 1.3. Small deviations from flatness produce substantial changes in the Hubble constant, with the product h[tot]^5
remaining roughly constant [Tegmark et al. (2004b)].
Suppose we were to show table 1 to our randomly chosen IAU members and again ask them the value of the Hubble constant (and its uncertainty). Some would argue that the results for the very-nearly
flat case are so very-nearly flat that it is reasonable to adopt perfect flatness as a working model. We observe that | log[tot]| < 0.10, when it might have been of order 100 (or perhaps 3 if one
admits anthropic reasoning). But others would be reluctant to take perfect flatness for granted. The very sensitivity of the Hubble constant to that assumption would be an argument against adopting
it. The present author is, himself, ambivalent. | {"url":"http://ned.ipac.caltech.edu/level5/March04/Schechter/Schechter1.html","timestamp":"2014-04-17T06:47:57Z","content_type":null,"content_length":"7209","record_id":"<urn:uuid:5161fe42-452f-4656-a129-98938e6722df>","cc-path":"CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00655-ip-10-147-4-33.ec2.internal.warc.gz"} |
Calculus Rotations Help!!!!
May 20th 2009, 08:25 AM #1
May 2009
Hello, I'm am really suck on these problems. If you're really smart I could use your help right now!
Use the disk method to find the volume of the solid:
y=2x-1, x=0, y=x^2 rotated about line y=3
Use the disk and shell method to solve this rotation:
y=-x^2+16, Y=16, X=4 about line x=2
Use either method to solve this rotation:
y=cos(x), y=sin(x), x=pi/4, x=5pi/4, about line x=5pi/4
Please help. I've been working on then for hours and have some equations but I don' know if they are right!
Hello, I'm am really suck on these problems. If you're really smart I could use your help right now!
Use the disk method to find the volume of the solid:
y=2x-1, x=0, y=x^2 rotated about line y=3
Use the disk and shell method to solve this rotation:
y=-x^2+16, Y=16, X=4 about line x=2
Use either method to solve this rotation:
y=cos(x), y=sin(x), x=pi/4, x=5pi/4, about line x=5pi/4
Please help. I've been working on then for hours and have some equations but I don' know if they are right!
I'm going to do the last one only and leave the other two for others.
Ok, here's what I think the integral should be:
$2piINTEG: (sinx-cosx)(5pi/4 -x) dx$ with limits from x=pi/4 to x=5pi/4
thanks for the help! I solved the others in class already. It just took a little longer than I thought to get to the solutions.
May 20th 2009, 09:03 AM #2
Dec 2008
May 20th 2009, 01:52 PM #3
May 2009 | {"url":"http://mathhelpforum.com/calculus/89795-calculus-rotations-help.html","timestamp":"2014-04-17T14:30:24Z","content_type":null,"content_length":"34604","record_id":"<urn:uuid:9838b6f7-ddd9-48c5-8c9c-aaebe2dfce06>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00312-ip-10-147-4-33.ec2.internal.warc.gz"} |
Calculate distance on a wrapped map
March 12th 2009, 05:17 PM
Calculate distance on a wrapped map
So I'm writing code that calculates distance between two cities for civilization 4 (modding, it's a hobby of mine). That's simple enough until you try to figure out how to deal with the wrapped
map. I've taken alot of math, been 5 years or so but got through Calc III. But I just can't figure this one out. I've tried googling the answer and got nothing.
What would be a good formula for calculating distance on a wrapped map (ie in a standard cartesian plane if X(max) + 1 = X(min) --also X(min) will always be 1, but X(max) will change depending on
the size of the map chosen by the player at game start-the calculation is done on a static map though once it's applied)? All x and y values in this plane are positive integers, not that that's
probably important. Also the ability of the x and y values to wrap is independent (can be changed on game setup), so it's best to analyize the x and y seperatly and just apply pythagoras later.
Though if that's impossible, I could just set up a bunch of if statements in the code.
March 13th 2009, 03:05 AM
For anyone interested this is how it was solved:
deltaX = abs(pCity.getX() - capital.getX())
if map.isWrapX():
deltaX = min(deltaX, map.getGridWidth() - deltaX)
deltaY = abs(pCity.getY() - capital.getY())
if map.isWrapX():
deltaY = min(deltaY, map.getGridWidth() - deltaY)
cityDistRaw = ( deltaX**2 + deltaY**2 )**0.5
For those who don't speak code (though this should be pretty easy to decifer anyway with math)
dX(i) = |P(x) - C(x)|
For a wrapped map
dX(f) = X(max) - dX(i)
if dX(f) < dX(i) then dX(f) = dX(i)
Repeat for Y component & then apply pythagoras to deltaX and deltaY. | {"url":"http://mathhelpforum.com/differential-geometry/78444-calculate-distance-wrapped-map-print.html","timestamp":"2014-04-19T07:31:57Z","content_type":null,"content_length":"5678","record_id":"<urn:uuid:2cbaccda-a03e-443b-8e47-058d314766b2>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00084-ip-10-147-4-33.ec2.internal.warc.gz"} |
Programmer Competency Matrix - 32 attributes to evaluate programmers
Programmer Competency Matrix – 32 attributes to evaluate programmers
Having worked with programmers with an extreme variance in skills, I sometimes get the feeling that there is an big lack of good programmers but when I thought about it a little more I realized that
it’s not very clear cut, some of the programmers have strong areas and if you confine the tasks into their strong areas then they tend to deliver well. So I started thinking about all the lines on
which we can evaluate a programmer, here’s what I have so far…
Programmer Competency Matrix (the table is too big to fit on this blog post and needs a whole page of it’s own)
After having spent a whole afternoon on this I realize that even this is not comprehensive, this matrix is more biased towards non-visual programmers, so a big majority of web devs will not be able
to relate well to this matrix, but I am tired and will come back to this at a later time.
Well done excellent table and I will use this to gauge my skills as a developer so far I’m primarily in Level 1 and Level 2. :(. But this is definitely a step in the right direction so now I know
what I have to learn and aim for. Even though this isn’t official it’s good in my book.
Great job! This might come in handy to rate programmers after a job interview. Personally I still have some catchup to do
Excellent way of breaking down observations. I am not sure how other are, but I am all over the levels randing from 1 to 3 on all aspects. Atleast I’m not at Level 0 – phew!
Brilliant table. Covers the bases nicely. Nice way of summing up the important differences between the stages of expertise.
Hi, thank you for the matrix its a useful tool to gauge strengths and weaknesses. I was wondering how much a programmer that is at 95% on Level2 and 50% level 3 would on average be expected to earn
in a profitable and innovative company in the United States.
I especially like that the table focuses on areas of competence that are relevant no matter if you work on GUI, middleware, database, batchjobs, soft realtime, mobile etc.
The “automated testing” row could be improved by level 2 being “used TDD regularly” and level 3 being “knows why TDD is a most often a bad idea”
Also I missed coverage of ability to work with customers, bosses, and specialists in other fields than software.
Great work. I will definately use it.
I found some errors in “languages exposed to”:
– SQL does not strike me as a declarative language.
– Imperative and object oriented are orthogonal. You can have languages that are both, neither and just one of the two.
– Java is an imperative, object oriented language.
– Haskell is not imperative, but object oriented.
– xslt is not imperative and not object oriented.
– Most older versions of Basic are imperative and not object oriented.
AFAIK the most common language classes (that are usually mutually exclusive) are imperative, declarative and functional. Some would add logic programming, but thats usually contained in the class of
declarative languages.
Hi, sorry to post again, but just wondering if you would post what your estimate would be on how much a programmer would be expected to earn at the level indicated above, which would be a
conservative estimate (at least that much experience, I know it would be dependant upon geography, which I could use a service like payscale to equate to other areas). Thank you, Glad to see your
site is back online….
two minor suggestions:
link back to this page from the matrix
add “Array (APL, J, MATLAB)” to the log(n) column of the two language rows
The guy that wrote the matrix hasn’t enough experience to be able to determine what people need to know. When I compare my 28 years of professional experience to his 7 or so years, I find that from a
managerial standpoint (I’m a senior consultant and technical manager for a Fortune 400 company which means I review code, write designs, and mentor developers) there are a lot of things missing from
that table, and a lot of things that matter not at all.
(For the record, I’ve written code for 737 flight trainers, show control software in Las Vegas hotels, and pioneered hardware and software products that changed the face of the travel industry. I
even wrote a few games back in the late 80′s.)
What the writer fails to perceive is that it isn’t what you know, it’s what you do with the knowledge. We’ve certainly had to fire programmers that would have ranked in the O log(n) columns for many
categories simply because they couldn’t get anything done. (I work for a rather large multi-national corporation that produces both hardware and software.) We gave one guy a project and three weeks
later there wasn’t one line of code. The analysis and design was done, mind you–he was still cogitating on the relative merits of for() vs. while() vs. do-while(). In a world where payment rightfully
comes from results, getting results is important–as opposed to whether or not one has “tried out Bzr/Mercurial/Darcs/Git.”
We also have a lot of bright third-column guys that simply spend too much time screwing off instead of getting work done. Work ethic isn’t discussed at all in the matrix, and it’s actually more
important than whether or not someone compiles from the IDE or on the command line.
The matrix also doesn’t really address communications at all–it just sort of takes a lazy swipe at it and then goes back to lazy crap like “who knows the most keyboard shortcuts.” Honestly, the whole
thing is so grade six that it’s embarrassing.
As an exercise in mental masturbation, I give it a B+. As a reflection of real-world skill requirements, I give it a C-, if not a D+. The author, 20 years from now, will find it an embarrassment and
hope no one downloaded and kept a copy.
Hi James,
Thanks for your inputs, This post was written a year back and I already think that some of the stuff in here is obsolete or needs revision. I can only write based on the experience that I have right
now and hopefully a few years from now I can come up with a better matrix that covers more of the “getting things done” aspects that you talk about (which I agree are very important as well)
This appeared to be a Microsoft Technologies Competency Matrix. Reading SICP or the Little Schemer does not get you to level 3 in the other crowd. SICP is introductory material.
- Is able to effectively communicate with peers – is a Level 2 skill?
- Writing a framework means little; Most “Architects” i have seen have written one. Mostly unusable.
- File has license header, summary, well commented, consistent white space usage. The file should look beautiful?
- Scott Hanselman’s Power Tools list?
- ….
I guess the list is partly valid within the Microsoft-Technology + IT Services context.
I like the idea of the competency matrix and I think it could be very useful. There are a few topics I would add to the matrix somewhere. I know some of these are more from a software project
management perspective, and I don’t know if you had intended to capture that in your matrix. However, since some people are going to want to use this matrix in order to gain the skills to climb up
the ladder, adding software project management to the list of skills could be useful.
Software process, lifecycles, process improvement (e.g. agile, XP, SCRUM, RUP, TSP, CMMI). Software architecture documentation (e.g. UML notations, component-and-connector diagrams). Software
architecture and design (e.g. design patterns, architectural styles). Formal methods (OCL, Z, Petri Nets, rate monotonic analysis). Static analysis. Software project estimation (use case points,
function points, COCOMO). Requirements documentation (use cases, user stories, quality attributes and quality attribute scenarios, paper prototypes). Project planning and management (work breakdown
structures, earned value analysis). Risk management.
James Hawk III, thank you for your comment on this!
Regarding James Hawk III’s comments – I think these are a unnecessary viscous (and probably shows someones belief in their own self importance). I’ve been developing real-time embedded systems for
both military and commercial systems since the mid-1980′s (and no I haven’t “changed the face” of those). Yes I agree that maybe the chart is based more on knowledge rather than competency
(competency – the proven ability to use knowledge, skills and other abilities to perform a function against a given standard in work), but this is very common confusion (see Microsoft MCSE etc.). The
other thing that could be improved is clarifying the target domain (many of the things there aren’t applicable to real-time embedded systems) – e.g. what type of applications. I worry about James’
managerial style if doesn’t understand the basics of constructive feedback – let’s see your attempt then? Overall it’s a good first attempt and gives a useful vehicle for discussion (as per Sam’s
I don’t really agree with the way this is setup. You should really get down to the essentials of programming. You mention books like Code Complete, yet you fail to take it’s advice (traits of a good
programmer). Traits are way more important than tool or even language knowledge. Communication is one. Depending on the situation (programming games, space shuttles, web apps, bookkeeping) influences
what you need to know. Knowing a gazillion languages doesn’t help in a project if you’re working in only one. Patterns should emerge out of good design and common sense, not memorized etc. And I am
not putting it down because I don’t understand the matrix, I do. Most modern languages used nowadays are imperative OO (Java, PHP, .NET) so it would help to master at least that.
The problem with this whole thing is the lack of #ux (user experience and usability). All the points made are moot if you’re building /the wrong product/.
Some other things to point out:
DVCS: lacking for the level of knowledge needed at a “log n” level. Rebasing with GIT or scripting something in SVN to act like GIT’s bisect functionality.
also missing. ability to work with branch-per-feature, merging strategies, VCS hooks and custom attributes.
architecture part is missing. No mention of DDD or description of where OO came from and why
IoC, DI, the rest of SOLID. AOP
where to use and where not to use
where and why RAD is good / bad
IDE is very shortsighted making custom macros look like a holy grail. Refactoring tools like IntelliJ IDEA, Resharper, Refactor Pro, Code Rush?
Just initial thoughts, I tweeted the same thing but may have left something out.
Otherwise, a great place to start discussion.
That is a helpful programmer competency matrix. I would like to set this up as a goal to move as close to level 2 as possible in about a year. Perhaps mentors should show this matrix to beginners and
give them a rough idea about what they should do to become better programmers.
I found a similar matrix created by an organization. You can find it on my blog at http://kevinrodrigues.com/blog/2009/12/28/what-skills-should-a-good-developer-possess/
The Database knowledge row is woefully inadequate. Programmers a step above 2^n, your n^2 level, generally have no idea what ACID stands for, let alone anything further up the chart. n^2 should be
“Basic competence in SQL. Can design tables for simple task. Has heard of Indexes.” Then shift the exist n^2 and n right one place. Parts of level logn reads more like a DBA job description than a
programmer competence description and doesn’t belong on the chart.
Similarly, you 2^n blob level is wrong too. Most low-competency programmers have no idea there are programming blogs. I know some folks who hit logn status in parts of your matrix who would respond
“Really? There are programming blogs?” if asked what they read.
This is the single most motivating thing I’ve seen in – well I can’t even remember how much. Thank you.
The only point I’d strongly disagree with is deep API usage knowledge. The whole point of APIs is to just take them as you go, when you need them, in a no-brainer way – hence all the authors’ efforts
on naming, parameters, functionality description, local organization into packages… Browsing the source of the Java class library is much more worthy of one’s time than memorizing a bunch of method | {"url":"http://sijinjoseph.com/2008/04/30/programmer-competency-matrix/","timestamp":"2014-04-19T09:34:40Z","content_type":null,"content_length":"59423","record_id":"<urn:uuid:252c3d63-6516-4d21-9b10-aed72f56b2d1>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00346-ip-10-147-4-33.ec2.internal.warc.gz"} |
A frustrating cohomology class on the moduli of abelian surfaces
up vote 5 down vote favorite
Here's a very frustrating question that I have been stuck on for some time. I believe that my question could fit in a general framework of what happens when you restrict $L^2$-cohomology classes on a
Shimura variety to a sub-Shimura variety. However I formulate the question for the special case I am interested in.
Let $A_2$ be the moduli stack of principally polarized abelian surfaces. To the irreducible finite dimensional representation of $\mathrm{Sp}(4)$ of highest weight $a \geq b \geq 0$ we attach a a
local system $V_{a,b}$ on $A_2$.
Suppose $(a,b) \neq (0,0)$. One can prove that $H^4_c(A_2,V_{a,b})$ vanishes unless $a=b$ is even, in which case $H^4_c(A_2,V_{2k,2k})$ is pure of Tate type and of the same dimension as the space of
cusp forms of weight $4k+4$ for $\mathrm{SL}(2,\mathbf Z)$. The map $H^4_c \to H^4_{(2)}$ to the $L^2$-cohomology is an isomorphism. In terms of automorphic representations, these cohomology classes
can be described as follows: for any level 1 cusp form $\pi$ on $\mathrm{GL}(2,\mathbf A)$ of weight $4k+4$ we consider the unique irreducible quotient of $$ \mathrm{Ind}_{P(\mathbf A)}^{\mathrm{GSp}
(4,\mathbf A)} \left( \vert \cdot \vert^{1/2} \pi \otimes \vert \cdot \vert^{-1/2} \right)$$ where $P$ denotes the Siegel parabolic subgroup (whose Levi factor is $\mathrm{GL}(2) \times \mathrm{GL}
(1)$); this is a discrete automorphic representation for $\mathrm{GSp}(4)$ which contributes a Tate type class to the $L^2$-cohomology in degrees $2$ and $4$.
There is a map $\mathrm{Sym}^2(A_1) \hookrightarrow A_2$ given by taking a pair of elliptic curves to their product. We can also restrict $V_{a,b}$ to $\mathrm{Sym}^2(A_1)$. By determining the
branching formula for $\mathrm{SL}(2)^2 \rtimes S_2 \subset \mathrm{Sp}(4)$ we find that the trivial local system occurs as a summand in the restriction of $V_{a,b}$ to $\mathrm{Sym}^2(A_1)$ if and
only if $a=b$ is even, in which case it appears with multiplicity $1$. So $H^4_c(\mathrm{Sym}^2(A_1),V_{2k,2k})$ is also pure of Tate type but $1$-dimensional. Again we could think about $L^
2$-cohomology and it would not make a difference.
MAIN QUESTION: Is the restriction map $H^4_c(A_2,V_{2k,2k}) \to H^4_c(\mathrm{Sym}^2(A_1),V_{2k,2k})$ nonzero for $k \geq 2$?
Any ideas or pointers at all would be appreciated. I am very ignorant about automorphic representations, Shimura varieties etc. and I am naively hoping that there exists some general method for
answering question of this form.
This question arose from the paper http://arxiv.org/abs/1210.5761 . A positive answer would imply that all even cohomology of $\mathcal{\overline{M}}_{2,n}$ is tautological for $n < 20$, and that the
Gorenstein conjecture fails on $\mathcal{\overline{M}}_{2,20}$.
add comment
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MathGroup Archive: July 2005 [00299]
[Date Index] [Thread Index] [Author Index]
Re: positive square root
• To: mathgroup at smc.vnet.net
• Subject: [mg58700] Re: positive square root
• From: Peter Pein <petsie at dordos.net>
• Date: Fri, 15 Jul 2005 03:02:15 -0400 (EDT)
• References: <db57ne$4no$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
paulvonhippel at yahoo schrieb:
> I work in a world where the square root is always a positive number.
> But Mathematica allows for the possibility of negative square roots.
> Two questions arise:
> (1) Is there a way to tell Mathematica that I'm only interested in
> positive square roots?
> (2) My current solution is to use, e.g., Abs[Sqrt[z]]. But when
> Mathematica echoes this, it puts the Abs function *under* the radical,
> so it looks like Sqrt[Abs[z]]. Is this a bug in the display?
Hi Paul,
I don't think so. It is just in general easier, to evaluate Sqrt[Abs[z]]
than Abs[Sqrt[z]]:
Assuming[(x | y) \[Element] Reals,
{Abs[ComplexExpand[Sqrt[x + I*y]]],(*sqrt first*)
ComplexExpand[Sqrt[Abs[x + I*y]]]}]](*abs first*)
{(x^2 + y^2)^(1/4)*
Abs[Cos[(1/2)*ArcTan[x, y]] + I*Sin[(1/2)*ArcTan[x, y]]],
x^2 + y^2)^(1/4)}
These are of course equivalent:
SameQ @@ Simplify[ComplexExpand[%]]
Peter Pein | {"url":"http://forums.wolfram.com/mathgroup/archive/2005/Jul/msg00299.html","timestamp":"2014-04-16T13:27:52Z","content_type":null,"content_length":"35194","record_id":"<urn:uuid:a75fc08c-efda-4c5e-b8c7-88178a52ebbc>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00094-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Topic: Cech Stone Compactification
Replies: 7 Last Post: Jan 2, 2013 3:43 AM
Messages: [ Previous | Next ]
A N Niel Re: Cech Stone Compactification
Posted: Dec 31, 2012 8:35 AM
Posts: 2,243
Registered: 12/7/04 In article <Pine.NEB.4.64.1212310124010.26424@panix2.panix.com>,
William Elliot <marsh@panix.com> wrote:
> Would somebody elucidate what Wikipedia was saying about
> Stone-Cech compactification? It doesn't make sense for
> isn't a compactification an embedding into an compact space.
> Some authors add the assumption that the starting space be Tychonoff
> (or even locally compact Hausdorff), for the following reasons:
> * The map from X to its image in bX is a homeomorphism if and only
> if X is Tychonoff.
> * The map from X to its image in bX is a homeomorphism to an open
> subspace if and only if X is locally compact Hausdorff.
> The Stone-Cech construction can be performed for more general spaces
> X, but the map X -> bX need not be a homeomorphism to the image of X
> (and sometimes is not even injective).
So, in those more general spaces, the *construction* can still be
carried out. But the map is not a homeomorphism. So the result of the
construction is not a "compactification" in your sense.
Date Subject Author
12/31/12 Re: Cech Stone Compactification A N Niel
12/31/12 Re: Cech Stone Compactification David C. Ullrich
12/31/12 Re: Cech Stone Compactification William Elliot
12/31/12 Re: Cech Stone Compactification Herman Rubin
1/1/13 Re: Cech Stone Compactification William Elliot
1/1/13 Re: Cech Stone Compactification David Hartley
1/2/13 Re: Cech Stone Compactification William Elliot | {"url":"http://mathforum.org/kb/thread.jspa?threadID=2423261&messageID=7945756","timestamp":"2014-04-17T10:50:34Z","content_type":null,"content_length":"24000","record_id":"<urn:uuid:a0c18205-ee71-4d98-839b-3ef4383e1669>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00614-ip-10-147-4-33.ec2.internal.warc.gz"} |
[VarianceForecast, H] = ugarchpred(U, Kappa, Alpha, Beta,
[VarianceForecast, H] = ugarchpred(U, Kappa, Alpha, Beta, NumPeriods) forecasts the conditional variance of univariate GARCH(P,Q) processes.
VarianceForecast is a number of periods (NUMPERIODS)-by-1 vector of the minimum mean-square error forecast of the conditional variance of the innovations time series vector U (that is, ɛ[t]). The
first element contains the 1-period-ahead forecast, the second element contains the 2-period-ahead forecast, and so on. Thus, if a forecast horizon greater than 1 is specified (NUMPERIODS > 1), the
forecasts of all intermediate horizons are returned as well. In this case, the last element contains the variance forecast of the specified horizon, NumPeriods from the most recent observation in U.
H is a vector of the conditional variances (σ[t]^2) corresponding to the innovations vector U. It is inferred from the innovations U, and is a reconstruction of the "past" conditional variances,
whereas the VarianceForecast output represents the projection of conditional variances into the "future." This sequence is based on setting pre-sample values of σ[t]^2 to the unconditional variance
of the {ɛ[t]} process. H is a single column vector of the same length as the input innovations vector U.
The time-conditional variance, , of a GARCH(P,Q) process is modeled as
where α represents the argument Alpha, β represents Beta, and the GARCH(P,Q) coefficients {Κ, α, β} are subject to the following constraints.
Note that U is a vector of residuals or innovations (ɛ[t]) of an econometric model, representing a mean-zero, discrete-time stochastic process.
Although is generated using the equation above, ɛ[t] and are related as
where is an independent, identically distributed (iid) sequence ~ N(0,1).
│ Note ugarchpred corresponds generally to the Econometrics Toolbox™ function garchpred. The Econometrics Toolbox software provides a comprehensive and integrated computing environment for the │
│ analysis of volatility in time series. For information, see the Econometrics Toolbox documentation or the financial products Web page at http://www.mathworks.com/products/finprod/. │ | {"url":"http://www.mathworks.co.uk/help/finance/ugarchpred.html?nocookie=true","timestamp":"2014-04-24T07:30:07Z","content_type":null,"content_length":"46277","record_id":"<urn:uuid:f7dcb266-8834-4862-b622-28ef434137a9>","cc-path":"CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00503-ip-10-147-4-33.ec2.internal.warc.gz"} |
A181874 - OEIS
A181874 Minute hand closest to hour hand on analog quartz clock. Best approximation for seconds. 1
0, 27, 55, 22, 49, 16, 44, 11, 38, 5, 33 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2
COMMENTS At which a.m. times h:m:s is the minute hand closest to the hour hand on an analog quartz clock (discrete seconds)? For an analog clock with continuous seconds this is the overlap problem
nr. 43 of the quoted Loyd/Gardner book where also the solution is given (pp. 41-2, solution pp. 180-1 in the German version). See A183032.
REFERENCES Sam Loyd, Mathematische Raetsel und Spiele, ausgewaehlt und herausgegeben von Martin Gardner, Dumont, Koeln, 1978, 3. Auflage 1997.
Sam Loyd, Mathematical puzzles, selected and edited by Martin Gardner, Dover, 1959.
LINKS Table of n, a(n) for n=0..10.
FORMULA a(n) gives the second for the (a.m.) hour h=n = 0,1,2,...,10, when the minute hand is closest to the hour hand on an analog quartz clock (discrete seconds), provided the minute is A178181
a(n)= round((300/11)*n) (mod 60), n=0..10. See the solution in the Loyd book with (27+3/11)s = 300/11 s.
EXAMPLE The eleven times are:
00:00:00, 01:05:27, 02:10:55, 03:16:22,
04:21:49, 05:27:16, 06:32:44, 07:38:11,
08:43:38, 09:49:05, 10:54:33.
The next time would be 12:00:00
CROSSREFS Cf.: A178181, A183032, A183033.
Sequence in context: A033903 A175806 A232922 * A042444 A042442 A042440
Adjacent sequences: A181871 A181872 A181873 * A181875 A181876 A181877
KEYWORD nonn,fini,easy
AUTHOR Wolfdieter Lang, Jan 21 2011
STATUS approved | {"url":"http://oeis.org/A181874","timestamp":"2014-04-19T23:46:58Z","content_type":null,"content_length":"16650","record_id":"<urn:uuid:a23e87c6-1715-4ac9-b251-17c455342a04>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00380-ip-10-147-4-33.ec2.internal.warc.gz"} |
How to calculate pad deformation
What was wrong with the link I offered?
The Wiki formula is not the one you want directly.
Do you understand what the symbols mean in it?
I understand that it's sth. like calculating efective elastic modulus of two serial springs.
[tex]\frac {1}{E} = \frac{1}{E_1} + \frac{1}{E_2}[/tex]
E is elastic modulus.
It's because we can say it's like two springs. One is the object and second is the pad.
And it is almost all I know about it.
I know what υ is. But I don't know how we get this. | {"url":"http://www.physicsforums.com/showthread.php?p=4214187","timestamp":"2014-04-16T10:21:06Z","content_type":null,"content_length":"63567","record_id":"<urn:uuid:a89b45e9-59e8-45b9-849d-fd8288de8003>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00319-ip-10-147-4-33.ec2.internal.warc.gz"} |
Generalized Tractability for Multivariate Problems: Part II: Linear Tensor Product Problems, Linear Information, and Unrestricted Tractability
Generalized Tractability for Multivariate Problems: Part II: Linear Tensor Product Problems, Linear Information, and Unrestricted Tractability
Permanent URL:
Part Number:
Department of Computer Science, Columbia University
Publisher Location:
New York
\usepackage{amssymb} \begin{document} We continue the study of generalized tractability initiated in our previous paper ``Generalized tractability for multivariate problems, Part I: Linear tensor
product problems and linear information'', J. Complexity, 23, 262-295 (2007). We study linear tensor product problems for which we can compute linear information which is given by arbitrary
continuous linear functionals. We want to approximate an operator $S_d$ given as the $d$-fold tensor product of a compact linear operator $S_1$ for $d=1,2,\dots\,$, with $\|S_1\|=1$ and $S_1$ has
at least two positive singular values. Let $n(\varepsilon,S_d)$ be the minimal number of information evaluations needed to approximate $S_d$ to within $\varepsilon\in[0,1]$. We study \emph
{generalized tractability} by verifying when $n(\varepsilon,S_d)$ can be bounded by a multiple of a power of $T(\varepsilon^{-1},d)$ for all $(\varepsilon^{-1},d)\in\Omega \subseteq[1,\infty)\
times \mathbb{N}$. Here, $T$ is a \emph{tractability} function which is non-decreasing in both variables and grows slower than exponentially to infinity. We study the \emph{exponent of
tractability} which is the smallest power of $T(\varepsilon^{-1},d)$ whose multiple bounds $n(\varepsilon,S_d)$. We also study \emph{weak tractability}, i.e., when $\lim_{\varepsilon^{-1}+d\to\
infty,(\varepsilon^{-1},d)\in\Omega} \ln\,n(\varepsilon,S_d)/(\varepsilon^{-1}+d)=0$. In our previous paper, we studied generalized tractability for proper subsets $\Omega$ of $[1,\infty)\times\
mathbb{N}$, whereas in this paper we take the unrestricted domain $\Omega^{\rm unr}=[1,\infty)\times\mathbb{N}$. We consider the three cases for which we have only finitely many positive singular
values of $S_1$, or they decay exponentially or polynomially fast. Weak tractability holds for these three cases, and for all linear tensor product problems for which the singular values of $S_1$
decay slightly faster that logarithmically. We provide necessary and sufficient conditions on the function~$T$ such that generalized tractability holds. These conditions are obtained in terms of
the singular values of $S_1$ and mostly limiting properties of $T$. The tractability conditions tell us how fast $T$ must go to infinity. It is known that $T$ must go to infinity faster than
polynomially. We show that generalized tractability is obtained for $T(x,y)=x^{1+\ln\,y}$. We also study tractability functions $T$ of product form, $T(x,y) =f_1(x)f_2(x)$. Assume that $a_i=\
liminf_{x\to\infty}(\ln\,\ln f_i(x))/(\ln\,\ln\,x)$ is finite for $i=1,2$. Then generalized tractability takes place iff $$a_i>1 \ \ \mbox{and}\ \ (a_1-1)(a_2-1)\ge1,$$ and if $(a_1-1)(a_2-1)=1$
then we need to assume one more condition given in the paper. If $(a_1-1)(a_2-1)>1$ then the exponent of tractability is zero, and if $(a_1-1)(a_2-1)=1$ then the exponent of tractability is
finite. It is interesting to add that for $T$ being of the product form, the tractability conditions as well as the exponent of tractability depend only on the second singular eigenvalue of $S_1$
and they do \emph{not} depend on the rate of their decay. Finally, we compare the results obtained in this paper for the unrestricted domain $\Omega^{\rm unr}$ with the results from our previous
paper obtained for the restricted domain $\Omega^{\rm res}= [1,\infty)\times\{1,2,\dots,d^*\}\,\cup\,[1,\varepsilon_0^{-1})\times\mathbb{N}$ with $d^*\ge1$ and $\varepsilon_0\in(0,1)$. In
general, the tractability results are quite different. We may have generalized tractability for the restricted domain and no generalized tractability for the unrestricted domain which is the
case, for instance, for polynomial tractability $T(x,y)=xy$. We may also have generalized tractability for both domains with different or with the same exponents of tractability. \end{document}
Item views: | {"url":"http://academiccommons.columbia.edu/catalog/ac%3A110791","timestamp":"2014-04-21T07:15:08Z","content_type":null,"content_length":"23452","record_id":"<urn:uuid:ca6f1dfe-55eb-40c6-a603-f99e1c16cbe0>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00091-ip-10-147-4-33.ec2.internal.warc.gz"} |
Dublin, CA Science Tutor
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SPEED: Precise and Efficient Static Estimation of Program Computational Complexity
This paper describes an inter-procedural technique for computing symbolic bounds on the number of statements a procedure executes in terms of its scalar inputs and user-defined quantitative functions
of input data-structures. Such computational complexity bounds for even simple programs are usually disjunctive, non-linear, and involve numerical properties of heaps. We address the challenges of
generating these bounds using two novel ideas. We introduce a proof methodology based on multiple counter instrumentation (each counter can be initialized and incremented at potentially multiple
program locations) that allows a given linear invariant generation tool to compute linear bounds individually on these counter variables. The bounds on these counters are then composed together to
generate total bounds that are non-linear and disjunctive. We also give an algorithm for automating this proof methodology. Our algorithm generates complexity bounds that are usually precise not only
in terms of the computational complexity, but also in terms of the constant factors. Next, we introduce the notion of user-defined quantitative functions that can be associated with abstract
data-structures, e.g., length of a list, height of a tree, etc. We show how to compute bounds in terms of these quantitative functions using a linear invariant generation tool that has support for
handling uninterpreted functions. We show application of this methodology to commonly used data-structures (namely lists, list of lists, trees, bit-vectors) using examples from Microsoft product
code. We observe that a few quantitative functions for each data-structure are usually sufficient to allow generation of symbolic complexity bounds of a variety of loops that iterate over these
data-structures, and that it is straightforward to define these quantitative functions. The combination of these techniques enables generation of precise computational complexity bounds for
real-world examples (drawn from Microsoft product code and C++ STL library code) for some of which it is non-trivial to even prove termination. Such automatically generated bounds are very useful for
early detection of egregious performance problems in large modular codebases that are constantly being changed by multiple developers who make heavy use of code written by others without a good
understanding of their implementation complexity. | {"url":"http://research.microsoft.com/en-us/um/people/sumitg/pubs/popl09_speed_abs.html","timestamp":"2014-04-17T02:51:13Z","content_type":null,"content_length":"2891","record_id":"<urn:uuid:9e5603ff-baac-4549-89e0-6059703b43eb>","cc-path":"CC-MAIN-2014-15/segments/1398223204388.12/warc/CC-MAIN-20140423032004-00009-ip-10-147-4-33.ec2.internal.warc.gz"} |
“The Bird of Time has but a little way
To flutter—and the Bird is on the Wing”
Which literary technique does the poet use in this excerpt?
A. simile
B. metaphor
C. apostrophe
D. assonance
“The Bird of Time has but a little way To flutter—and the Bird is on the Wing” Which literary technique does the poet use in this excerpt? A. simile B. metaphor C. apostrophe D. assonance
B. metaphor
Expert answered|ANGEEELICA|Points 1413|
Asked 2/18/2011 10:45:50 AM
0 Answers/Comments
Not a good answer? Get an answer now. (FREE)
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Ronitt Rubinfeld. Fast approximate PCPs
, 2003
"... Sensor networks promise viable solutions to many monitoring problems. However, the practical deployment of sensor networks faces many challenges imposed by real-world demands. Sensor nodes often
have limited computation and communication resources and battery power. Moreover, in many applications se ..."
Cited by 175 (11 self)
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Sensor networks promise viable solutions to many monitoring problems. However, the practical deployment of sensor networks faces many challenges imposed by real-world demands. Sensor nodes often have
limited computation and communication resources and battery power. Moreover, in many applications sensors are deployed in open environments, and hence are vulnerable to physical attacks, potentially
compromising the sensor's cryptographic keys. One of the basic and indispensable functionalities of sensor networks is the ability to answer queries over the data acquired by the sensors. The
resource constraints and security issues make designing mechanisms for information aggregation in large sensor networks particularly challenging.
- SIAM Journal on Computing , 2004
"... In this work we look back into the proof of the PCP Theorem, with the goal of finding new proofs that are “more combinatorial ” and arguably simpler. For that we introduce the notion of an
assignment tester, which is a strengthening of the standard PCP verifier, in the following sense. Given a state ..."
Cited by 24 (3 self)
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In this work we look back into the proof of the PCP Theorem, with the goal of finding new proofs that are “more combinatorial ” and arguably simpler. For that we introduce the notion of an assignment
tester, which is a strengthening of the standard PCP verifier, in the following sense. Given a statement and an alleged proof for it, while the PCP verifier checks correctness of the statement, the
assignment-tester checks correctness of the statement and the proof. This notion enables composition that is truly modular, i.e., one can compose two assignment-testers without any assumptions on how
they are constructed. A related notion called PCPs of Proximity was independently introduced in [Ben-Sasson et. al. STOC 04]. We provide a toolkit of (non-trivial) generic transformations on
assignment testers. These transformations may be interesting in their own right, and allow us to present the following two main results: 1. The first is a new proof of the PCP Theorem. This proof
relies on a rather weak assignment tester given as a “black box”. From this, we construct combinatorially the full PCP. An important component of this proof is a new combinatorial aggregation
technique (i.e., a new transformation that allows the verifier to read fewer, though possibly longer, “pieces ” of the proof). An implementation of the black-box tester can be obtained from the
algebraic proof techniques that already appear in [BFLS91, FGL + 91]. Obtaining a combinatorial implementation of this tester would give a purely combinatorial proof for the PCP theorem, which we
view as an interesting open problem. 2. Our second construction is a “standalone ” combinatorial construction showing NP ⊆ P CP [polylog, 1]. This implies, for example, that approximating max-SAT is
quasi-NPhard. This construction relies on a transformation that makes an assignment tester “oblivious”: so that the proof locations read are independent of the statement that is being proven. This
eliminates, in a rather surprising manner, the need for aggregation in a crucial point in the proof. 1
- Proc. of of ACM SenSys 2003 , 2003
"... ..."
, 2004
"... * The included papers are copyrights of the respective authors/universities and of ACM / IEEE respectively and are reproduced for academic purposes only. ..."
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* The included papers are copyrights of the respective authors/universities and of ACM / IEEE respectively and are reproduced for academic purposes only. | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=111420","timestamp":"2014-04-16T13:20:11Z","content_type":null,"content_length":"20012","record_id":"<urn:uuid:52524e74-6fe8-48f9-9974-c9ffb0633a50>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00224-ip-10-147-4-33.ec2.internal.warc.gz"} |
Tests for convergences
SUM(arctan(a_n) converges if and only if SUM(a_n) converges. a_n in this case is a sequence of positive numbers.
Last edited by Juancd08; July 9th 2009 at 06:39 PM.
[quote=Juancd08;337083]SUM(arctan(a_n) converges if and only if SUM(a_n) converges. preposition 1.Let arctan(a_n)=b_n => tan(b_n)=a_n 2.ıf any sequence is converges than its subsequences also
converge. Solution tan(b_n) is a subsequence of b_n. By the comparison princeple if b_n is converge then tan b_n also converge. Conversly now show that if tan(b_n) is converge than b_n converges.
Similiarly arctan(a_n) is subsequence of a_n then right handed also satisfies. then proof is done | {"url":"http://mathhelpforum.com/differential-geometry/94689-tests-convergences.html","timestamp":"2014-04-17T22:51:32Z","content_type":null,"content_length":"36257","record_id":"<urn:uuid:5cbbf527-955c-44c8-9493-2e22cc878501>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00050-ip-10-147-4-33.ec2.internal.warc.gz"} |
Physics Forums - View Single Post - Calc I problem regarding orthogonal curves and families of curves
It's really not so hard to solve for the intersection points. In fact, it's pretty easy. You've got y=ax/b. Put that into the quadratics. But I think the easiest way is to multiply your two
derivatives (a-2x)/(2y) and (2x)/(b-2y) and see if you can show it's (-1) just using the original equations. BTW, one of the intersection points is clearly (0,0). You might need to make a special
argument there. Sometimes this sort of question does take some cleverness to see an easy method. | {"url":"http://www.physicsforums.com/showpost.php?p=2897042&postcount=10","timestamp":"2014-04-16T07:34:01Z","content_type":null,"content_length":"7362","record_id":"<urn:uuid:ea8db4da-df86-492c-957a-15ca2d7587dd>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00107-ip-10-147-4-33.ec2.internal.warc.gz"} |
Some Gems from Physical Mathematics
Posted by: yanzhang | June 24, 2010
Some Gems from Physical Mathematics
It’s been a while since I gave cute problems, so here are a few with a theme: solving mathematical problems with physics (I’m not talking about the other way around, which is why I said “physical
mathematics” instead of “mathematical physics”).
• Take a triangle ABC. Prove that the point X inside the triangle which minimizes AX + BX + CX is a point such that the angles AXB, BXC, and CXA are all $120$ degrees.
• Consider an ellipse with ratio of the major to minor axes equal to $r > 1$. From either foci, start enlarging a circle until the circle is tangent to the ellipse from the inside. It is obvious
that the number of tangent points will be $1$ or $2$ and that this only depends on $r$. Given an $r$, figure out this count.
• Take a convex polyhedron and any point inside of it. To every face, drop a normal line from the point. Note that it is both possible to land inside the face or outside. Construct a polyhedron
where every such normal line drops outside of the corresponding face, or prove such a polyhedron cannot exist.
These are really delightful problems and among my all-time favorite – if you haven’t seen them before, I highly recommend thinking about them before the jump. =)
Here are the solutions:
• Take a table and drill $3$ holes so they form vertices of ABC. Now, take three strings and attach one end from all three at a single point X. At the other end of each of the three strings put a
weight of mass $m$. Put the weights through the holes. This system will seek to minimize potential energy, which happens when AX + BX + CX is minimized (if you don’t see this immediately,
consider the potential energy is just proportional to the negative of the sum of the three taut parts of the string under the table). However, this means the three forces from the three weights
must balance at X. Since the forces have the same tension (all $mg$), the forces must be at $120$ degrees from each other.
• It doesn’t matter what $r$ is – the answer is always $1$. This nonintuitive answer comes easily with some optical thinking: draw a tangent line $l$ to one of the tangent points $P$ and fire a ray
of light from the chosen focus to $P$. Note that because we have an ellipse, the reflection must go through the other focus. However, because the line is also tangent to the circle, the
reflection must also go through the center of the circle, which is the focus we started with! Since the reflection must go through both foci, the tangent must have been orthogonal to the major
axis, meaning we only had $1$ contact point.
• This polyhedron cannot exist. Suppose we put it on a flat table with bottom face $F$. Since by construction the normal line hits outside of $F$, the polyhedron must roll onto another face (yes,
or edge, but we can spent basically zero energy to push it onto a face using that edge instead, so the argument still works), but then the same thing happens and we get perpetual motion. As we
recall, perpetual motion cannot exist in our world =)
The first problem is fairly canonical – it has been around for a while and I might have seen it first in Titu Andresscu and Razvan Gelca’s “Mathematical Olympiad Challenges” back in the Olympiad
days. The second problem met me in middle school – it became one of my favorite problems ever to appear in a contest after I found it one day practicing old ARML/NYCIML problems (sadly I cannot
remember which contest this was from). I learned of the third problem from Tadashi Tokieda’s (seems like a really fun guy, by the way) article “Mechanical Ideas in Geometry” in an old Monthly.
The prompt for this post is that I recently rediscovered the first problem in Mark Levi’s gorgeous “The Mathematical Mechanic.” Though I was actually a bit sad since I’ve been working on a collection
much like his to share them in blog/book form at some point, I think it is really a great thing for the world that he put these problems out there in organized form, so everyone should go read that
book at some point and I’ll be happy vicariously through him. Besides, the second and third problems I gave are not in that book, so I can still share them here with you.
Have a nice summer, everyone,
Another nice one: on each face of a polyhedron, construct a vector perpendicular to the face with length proportional to the area of the face. Prove that the vectors sum to zero.
The polygonal version of this is pretty easy. This polyhedral version corresponds to the fact that a polyhedron filled with air and currently at rest will stay at rest, with regards to the air
pushing outward on each face.
By: A. Rex on June 24, 2010
at 2:27 PM
that’s def. a nice one. I may have used it if I didn’t already have a problem about polyhedra.
By: yanzhang on June 24, 2010
at 3:40 PM
Yan, we already discussed this, but here’s an alternate solution to the third question:
Suppose such a polyhedron and point exists. Among the altitudes from the point to the various faces, choose the shortest. Since the point is inside the polyhedron and the foot of the altitude is
outside, the altitude crosses some facet of the polyhedron. But then the altitude to this face is even shorter, which is a contradiction.
This solution (which is fundamentally equivalent to yours; we can think here of what happens to the potential energy of the arrangement as the polyhedron falls from face to face) is strongly
reminiscent of the “usual” solution of Sylvester’s Theorem. (I don’t know if HTML works here; if not, can you edit to make this look nice?) This raises the question of whether there is a “physical”
proof of Sylvester’s Theorem. Any takers?
By: JBL on June 26, 2010
at 3:17 AM
Here is yet another physical proof of the fist problem, which a physicist friend of mine came up with today:
Put 3 vertical poles at the points A,B and C and take another nonstationed vertical pole for the point X. Take now a rope, tie it at, say, A, pass it by X, then around B, then by X again, then around
C, then by X again and finally around A. Pull the rope tight at A. Since the length of the rope inside ABC is 2(AX +BX +CX) and since it’s tight, this length is minimal and X is at the desired
position. Now again as in the original proof, tension forces at X are equal in magnitude and sum to zero.
And a (not really)physical proof of the polyhedron problem: expand a spherical balloon centered at X inside the polyhedron until it touches the polyhedron. That point of contact is clearly inside and
is the foot of the perpendicular radius from X to the corresponding face.
By: Greta on June 27, 2010
at 2:18 AM
Wow, I really like the balloon thing
By: yanzhang on June 27, 2010
at 5:24 AM
Tanya Khovanova points me to the following, related to the third problem: http://mathworld.wolfram.com/UnistablePolyhedron.html
She remarks, “I believe that Conway suggested an idea how to make it with 18 faces” (rather than the 19 shown in the linked page).
By: JBL on June 29, 2010
at 1:17 AM
The first problem does not apply for obtuse triangles.
By: yan on August 27, 2010
at 12:42 AM
The relevant condition is actually that the triangle contain no angle larger than 120 degrees, not that the triangle be acute.
By: JBL on August 27, 2010
at 2:36 AM
You can fill a hollow polyhedron with sand, or float it in the ocean, and it will still reach an equilibrium. I wonder if this can be made into a puzzle.
By: Tk on October 11, 2010
at 12:47 AM
I’m a bit confused by the third one. I could see the solution given being correct if it were specified that the point selected must be the centroid (and indeed, given such information, the solution
given would be the first thing to pop into my head), but how can we get from “the existence of a point such that a line normal to each face through it will not pass through the associated face” to
“the polyhedron will tip over” when the point has no relationship to the polyhedron’s center of gravity?
By: David on April 1, 2011
at 8:26 AM
David, Yan seems to have left out a sentence from the explanation: you should take the entire mass of the polyhedron to be concentrated in the special point. (This makes it the center of gravity.)
By: JBL on April 1, 2011
at 2:00 PM
Posted in General | Tags: mathematics, physics | {"url":"http://concretenonsense.wordpress.com/2010/06/24/some-gems-from-physical-mathematics/","timestamp":"2014-04-21T14:43:18Z","content_type":null,"content_length":"79999","record_id":"<urn:uuid:932f85e5-66c5-451a-bb16-9bda0b4b305b>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00384-ip-10-147-4-33.ec2.internal.warc.gz"} |
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PURE_ONCE_ASM_REWRITE_TAC : thm list -> tactic
Rewrites a goal once, including the goal's assumptions as rewrites.
A set of rewrites generated from the assumptions of the goal and the supplied theorems is used to rewrite the term part of the goal, making only one pass over the goal. The basic tautologies are
not included as rewrite theorems. The order in which the given theorems are applied is an implementation matter and the user should not depend on any ordering. See GEN_REWRITE_TAC for more
information on rewriting tactics in general.
PURE_ONCE_ASM_REWRITE_TAC does not fail and does not diverge.
Manipulation of the goal by rewriting with its assumptions, in instances where rewriting with tautologies and recursive rewriting is undesirable.
ASM_REWRITE_TAC, GEN_REWRITE_TAC, ONCE_ASM_REWRITE_TAC, ONCE_REWRITE_TAC, PURE_ASM_REWRITE_TAC, PURE_ONCE_REWRITE_TAC, PURE_REWRITE_TAC, REWRITE_TAC, SUBST_ALL_TAC, SUBST1_TAC. | {"url":"http://www.cl.cam.ac.uk/~jrh13/hol-light/HTML/PURE_ONCE_ASM_REWRITE_TAC.html","timestamp":"2014-04-19T14:34:43Z","content_type":null,"content_length":"1994","record_id":"<urn:uuid:f6582c80-f33a-436c-b1f8-22a83e6be967>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00186-ip-10-147-4-33.ec2.internal.warc.gz"} |
Continuous dispersal on a discrete lattice
September 27, 2012
By Corey Chivers
Dispersal is a key process in many domains, and particularly in ecology. Individuals move in space, and this movement can be modelled as a random process following some kernel. The dispersal kernel
is simply a probability distribution describing the distance travelled in a given time frame. Since space is continuous, it is natural to use a continuous kernel. However, some modelling frameworks
are formulated on a lattice, or discrete array of patches.
So how can we implement a continuous kernel in discrete space?
As with many modelling situations, we could approach this in a number of ways. Here is the implementation that I can up with, and I welcome your suggestions, dear reader, for alternatives or
improvements to this approach.
1. Draw a random variate d from the dispersal kernel.
2. Draw a uniform random number θ between 0 and 2π, which we will use to choose a direction.
3. Calculate the relative location (in continuous space) of the dispersed individuals using some trig:
x = cos(θ) d
y = sin(θ) d
4. Determine the new location on the lattice for each individual by adding the relative x and y positions to the current location. Round these locations to the nearest integer and take the modulo of
this integer and the lattice size. This creates a torus out of the lattice such that the outer edges loop back on each other. If you remember the old Donkey Kong games, you can think of this like
how when you leave out the right side of the screen you enter from the left.
I implemented this approach in R as a function that takes in a population in a lattice, and returns a lattice with the dispersed population. The user can also specify which dispersal kernel to use.
Here is the result using a negative-exponential kernel on a 50×50 lattice.
Created by iterating over the dispersal function:
## General function to take in a lattice and disperse
## according to a user provided dispersal kernel
## Author: Corey Chivers
for(i in 1:lattice_size[1])
for(j in 1:lattice_size[2])
for(k in 1:N)
x_ind<-round(i+x[k]) %% lattice_size[1]
y_ind<-round(j+y[k]) %% lattice_size[2]
For comparison, I also ran the same population using a Gaussian kernel. I defined the parameters of both kernels to have a mean dispersal distance of 1.
Here is the result using a Gaussian kernel:
The resulting population after 35 time steps has a smaller range than when using the exponential kernel, highlighting the importance of the shape of the dispersal kernel for spreading populations
(remember that in both cases the average dispersal distance is the same).
Code for generating the plots:
############## Run and plot #######################
## Custom colour ramp
cus_col<-colorRampPalette(colors=colours, space = c("rgb", "Lab"),interpolate = c("linear", "spline"))
## Initialize population array
### Normal Kernel ###
for(i in 2:Time)
## Plot
png('normal_kern.png', width = 800, height = 800)
for(i in times)
main=paste("Time =",i))
mtext("Gaussian Kernel",outer=TRUE,cex=1.5)
### Exponential Kernel ###
for(i in 2:Time)
## Plot
png('exp_kern.png', width = 800, height = 800)
for(i in times)
main=paste("Time =",i))
mtext("Exponential Kernel",outer=TRUE,cex=1.5)
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Matches for: Author/Editor=(Stormer_Erling)
International Book Series of Mathematical Texts
2012; 148 pp; hardcover
ISBN-10: 973-87899-8-2
ISBN-13: 978-973-87899-8-2
List Price: US$36
Member Price: US$28.80
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This volume contains the proceedings of the 23rd International Conference on Operator Theory, held in Timişoara, Romania, from June 29 to July 4, 2010. It includes three survey articles on traces on
a \(C^*\)-algebra, amalgamated products of \(C^*\)-bundles, and compactness of composition operators, as well as ten papers containing original research on several topics: single operator theory, \(C
^*\)-algebras, moment problems, differential and integral operators, and complex function theory.
A publication of the Theta Foundation. Distributed worldwide, except in Romania, by the AMS.
Graduate students and research mathematicians interested in operator theory.
• A. M. Bikchentaev -- Characterization of traces on \(C^*\)-algebras: A survey
• E. Blanchard -- Amalgamated products of \(C^*\)-bundles
• E. Boasso -- The Drazin spectrum in Banach algebras
• B. E. Breckner and C. Varga -- On problems involving the weak Laplacian operator on the Sierpinski gasket
• S. Grigorian and A. Kuznetsova -- On a category of nuclear \(C^*\)-algebras
• L. Lemnete-Ninulescu -- Truncated trigonometric and Hausdorff moment problems for operators
• N. Lupa and M. Megan -- Rolewicz type theorems for nonuniform exponential stability of evolution operators on the half-line
• M. Megan, T. Ceauşu, and L. Biriş -- On some concepts of stability and instability for cocycles of linear operators in Banach spaces
• H. Queffélec -- A tentative comparison of compactness of composition operators on Hardy-Orlicz and Bergman-Orlicz spaces
• F. Rădulescu -- A universal, non-commutative \(C^\ast\)-algebra associated to the Hecke algebra of double cosets
• C. Stoica and M. Megan -- On nonuniform exponential dichotomy for linear skew-evolution semiflows in Banach spaces
• L. Suciu and N. Suciu -- On the asymptotic limit of a bicontraction
• A. Totoi -- Integral operators applied on classes of meromorphic functions defined by subordination and superordination | {"url":"http://cust-serv@ams.org/cgi-bin/bookstore/booksearch?fn=100&pg1=CN&s1=Stormer_Erling&arg9=Erling_St%F8rmer","timestamp":"2014-04-17T04:22:08Z","content_type":null,"content_length":"15974","record_id":"<urn:uuid:a059b05f-27ae-46d7-9446-a9ef3a61b2f5>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00327-ip-10-147-4-33.ec2.internal.warc.gz"} |
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hi, can anyone explain me how to get the integral: ∫((sin^3 x)/sqrt(cos x))dx
• one year ago
• one year ago
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well, you will need to take cos x as a variable, say 't', and then use chain rule to say that dx = dt/-sinx Then, you substitute for cos x and simplify and then integrate, and finally substitute
for 't' to get the answer.
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thanks a lot for your reply, but can you explain a little bit more de chain rule part?
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You first need to change sin^3x as follows: sin^3x =sin^2x sinx =(1-cos^2x)sinx Substitute this form into original integral and distribute cos^(-1/2) over (1-cos^2x) so you now have integral of
(cos^(-1/2)xsin x - cos^(3/2)xsinx )dx Now you can make the t substitution described earlier and you will have integral of t^(-1/2) -t^(3/2) dt which can be integrated by power rule.
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
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parenthesized braid operad
parenthesized braid operad
Algebraic theories
Algebras and modules
Higher algebras
Model category presentations
Geometry on formal duals of algebras
The parenthesized braid operad is an operad in Grpd modelled on the braid group.
Let $PaB$ be the category defined as follows:
• its set of objects is the free magma? on one generator, or equivalently the set of rooted binary tree?s.
• the set of morphisms between two objects $s,t$ is given by the braid group $B_n$ whenever $s$ and $t$ are words of the same legnth $n$, and is empty otherwise.
Then the collection $PaB$ of the $PaB_n$’s is a braided operad?. The composition
$\circ_i:PaB_n \times PaB_m \rightarrow PaB_{m+n-1}$
is given by replacing the $i$th strand of the first braid, by the second braid made very thin.
$PaB$ also have an obvious structure of a braided monoidal category. In fact:
Colored/ordered version
let $CPaB_n$ be the groupoid defined as follows:
• it set objects are parenthesized permutations of $\{1,\dots,n\}$, that is non-associative, non-commutative monomials on this set in which every letter appears exactly once.
• morphisms between two objects $s,t$ are braids connecting each letter in $s$ to the same letter in $t$. In other words, let $p:B_n\rightarrow S_n$ be the canonical projection from the braid group
to the symmetric group whose kernel is the pure braid group. Then, forgetting the parenthesization and viewing $s,t$ as permutations:
Then $CPaB$ is an (ordinary) operad, the operadic structure being the same as for the non-colored version.
A topological interpretation of $CPaB$ is as follows:
$PaB$ was originally defined in
Revised on September 16, 2012 15:06:26 by
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Converting from Polar to Rectangular/Parametric
May 3rd 2009, 08:20 PM #1
May 2009
Converting from Polar to Rectangular/Parametric
I've got a really neat graph out of r = 2 + cos(3theta/5), but I didn't know we had to have it in parametric form. Could someone tell me how to convert it? I know the whole y = rsin(theta) thing,
but I don't know how to deal with the fraction in the polar equation.
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Monarch Watch : Reading Room : Articles : Do Female Monarchs Show egg-Laying Preferences?
Do Female Monarchs Show Egg-Laying Preferences?
Christine Byerley, Orley R. Taylor, Stephanie Darnell
Department of Entomology, University of Kansas
Sandy Collins, West Junior High School, Lawrence, Kansas
In most of the Monarch s breeding areas, there are four to fifteen species of milkweeds, all of which are potential hosts for Monarch larvae. These plants differ in many aspects, such as growth
rates, toughness, hairiness, chemical make-up, and so on. This observation leads us to an interesting question about these varied plants: are they all equally attractive to Monarchs for oviposition
and feeding? Let's examine the oviposition question. How could we determine if females prefer one species over another? We should start with a hypothesis. The best approach is to start with a null
hypothesis (sometimes written H[o]). The null hypothesis states that there is no difference between milkweed species A and B. The purpose of designing a test around a null hypothesis is to avoid the
problem of unconsciously letting our expectations become part of our results. Our objective should be not to find that A is preferred more than B, if this is our expectation, but to establish in a
neutral and objective way the preference or lack of preference shown by the butterflies. In other words, if we have prior knowledge, or think we do, that species A is preferred to species B, we might
unintentionally obtain results which support this hypothesis. By suppressing our expectations, and doing a neutral test, we can let the butterflies tell us whether or not they show preferences. How
could we conduct such a test?
In the field it is very difficult to follow Monarchs and keep track of which milkweed species they use for egg laying. Even if we made such observations, it might not be possible to tell which
species they prefer. We would need to be able to determine if the ratio of contacts with the plants to eggs laid differed between or among milkweed species. However, in the laboratory we can provide
a choice. To test for oviposition preference, we will need mated female Monarchs (how many?), a large flight cage, artificial nectar, and at least two species of milkweed which we can use to test for
oviposition. If you set up such a cage and put in one species of milkweed and record the number of eggs per hour, you will notice that most of the egg laying occurs in the first 4-5 hours of light
each day. From this observation, it becomes clear that if we wish to test for preference, we should do so in the morning hours.
To conduct a test
What must we control for when we compare milkweed species? To start, milkweeds selected for this test should be of similar condition, i.e., both have new leaves (it would be inappropriate to compare
a plant of one species with old leaves to another with new leaves); both either with or without flowers (flowers may provide some type of stimulus - in fact, this is something we can test next: do
plants with flowers within the same species receive more or fewer eggs than those without flowers); both should be the same size; and so on. Should we control the position of plants within the flight
cages? There could be slight differences in the cage that could cause more eggs to be laid on the plant placed on one side of the cage than the other. Maybe the procedure should involve reversing the
positions of the plants; if we do this, how many times should we do so, and how many replicates do we need (i.e., how many pairs of plants should we compare)? Will two be enough? Four better? And how
long should the test interval be? 10 minutes? 20 minutes? 30 minutes? Should we use new plants on each test for each replicate? A few preliminary tests could be conducted to answer these procedural
Analyzing the results
After we have counted the eggs on each plant in each replicate, we can analyze the data to determine if there are differences between these plants in attractiveness. The choice of the approach here
depends on the level of sophistication of the students we are working with. We could simply total the results and if there were no substantial difference in the number of eggs laid on A and B, it
would be clear that the butterfly treats both species in a similar manner. If the students can handle relatively simple statistics, a chi-squared test or a t-test could be performed. These are
described in any beginning statistics book.
Interpretation of results
What are the possible interpretations of the results? The students can conclude either that: (1) there is no difference between A and B, or (2) there is a difference between A and B. Should the
results indicate a difference between these or other species, the task then becomes how to interpret this difference. Are there differences between the two species which are apparent to us? If the
students notice that A and B differ in leaf texture or hairiness, they might hypothesize that this may influence oviposition preference. Could the butterflies be using the same senses we do to select
plants? Or, do the butterflies use chemical information not available to us. If so, how do they perceive these signals and what types of chemicals are likely to be involved? Is there a way we could
make a less attractive species more attractive? By asking these questions the students are forced to make additional observations which can lead to questions and other hypotheses to test.
If you don't have more than one species of milkweed, but you have plants with flowers and plants with no flowers (remember, you can always cut flowers off of flowering stalks), you could conduct
tests to determine if the Monarchs show a difference in egg laying in the presence or absence of flowers. Similarly, we could ask whether females prefer to lay eggs on young (new) or old leaves. If
you can reject the null hypothesis, and there are more eggs laid on plants with or without flowers or new vs. old leaves, the students should be led to hypothesize why. This could lead to more
experiments, for example using filter paper extract of flowers, leaves, or both to help determine causes for these differences.
Add ons
Access to newly emerged Monarchs, milkweed plants and cages could be used by students to make observations on mating behavior, e.g., age at first mating, number of matings, time of day mating occurs,
etc. In addition, students could observe and record details about egg laying (oviposition) behavior, e.g. time of day, intervals between each oviposition, choice of oviposition sites, number of eggs
per day, and number of eggs per lifetime. These data could be graphed in a number of different ways to show lifetime patterns and differences between individuals. If students make observations of egg
laying they will notice that females lay most of their eggs on the underside of leaves. This leads to a why question that could be used to discuss the advantages and disadvantages of placing eggs
on upper vs. lower surfaces of leaves, and this discussion could lead to another experiment. What happens when we turn the plants upside down? Will the females continue to lay eggs on the lower
surface (now upper surface) as they did before? To conduct this test the students will have to figure out a way of suspending the plants upside down (without making a mess of everything). One
possible method is to use stem holders for roses which are available at most florists.
If students carry out the basic experiment outlined above and obtain results which indicate that females lay significantly more eggs on one type of plant, the next question is what is the underlying
basis for the difference? Do the females make choices before they land on the plants or after they have made contact? This question could be examined by making careful observations and records of
ratios of contact to egg laying.
Remember, to facilitate observations and data collecting, Monarchs can be marked to distinguish one from another - either with the use of tags or by using a permanent marking pen to give each
individual a number in the discal cell on the underside of the hindwing.
Number of eggs laid by female Monarchs (N=20) in 15 minute intervals.
(R) = plant positioned on right side of cage; (L) = plant positioned on left side of cage.
│ │Species A│Species B│
│Trial 1│ 102 (L)│ 46 (R)│
├───────┼─────────┼─────────┤ Is there a difference between A and B?
│Trial 2│ 52 (R)│ 114 (L)│
├───────┼─────────┼─────────┤ Is there a position effect in the cage?
│Trial 3│ 86 (L)│ 37 (R)│
│Trial 4│ 63 (R)│ 94 (L)│
Various statistical tests can be used to ask if there is a significant difference between two observations or sets of outcomes. The Chi Square (X^2) test is frequently used to determine whether
differences are likely to be due to chance, referred to as sampling variability (or sampling error), or likely to occur so rarely that they represent real or significant differences between two data
sets. If the differences between A and B are so small as to be due to chance, then there is no significant difference between them and we are unable to reject the null hypothesis. If, on the other
hand, the difference between A and B is significant at a probability of <.05% then we can reject the null hypothesis.
Steps for Performing X^2 Statistics
1) State the Null Hypothesis (H[0]) For example, female Monarchs show no egg-laying preferences and will lay similar numbers of eggs on milkweed species A and milkweed species B.
2) Calculate ECF (Expected Cell Frequency) based on marginal frequencies from the observed data.
Note: The expected data table with ECFs calculated.
3) Fill the expected data table with ECFs calculated.
4) Calculate X^2. The X^2 formula compares expected and observed cell frequencies to measure the difference between observed and expected values.
5) Calculate degrees of freedom (d.f.)
d.f. =( r-1)(c-1)
r = number of rows
c = number of columns
Degrees of freedom is the number of quantities that are unknown minus the number of independent equations linking these unknowns. For a contingency table, this is the number of cells that must be
filled in to determine all the cells. For example: a two by two table, the row and column marginals are known and the value in only one cell needs to be calculated to figure out all the other cells.
i.e. d.f. = (2-1)(2-1)= 1
6) Select a significance level. Most often .05 level is chosen. (Significance at the .05 level means that 5 out of 100 times the differences observed may have occurred by chance).
7) Look up the X^2 value on the Chi Square Distribution Table . Look at the value under the chosen significance level and the degrees of freedom calculated.
8) Check to see if the computed X^2 value is greater than the table value.
If larger, then H[0] is rejected and there is a significant difference.
If smaller, then failed to reject H[0] and there is no significant difference.
EXAMPLE 1: Is there an egg-laying preference between the two species of milkweed by the Monarch?
1) State the Null Hypothesis: There is no difference in egg laying preference between milkweed species A and B.
2) Calculate ECF, and 3) Fill the Expected Data Table
The values bolded are the marginal frequencies from the observed data (or known values).
Calculate the ECF for Species A - Trial 1 = (148 x 303) / 594
ECF = 75
Calculate the rest of the ECF s for Species A. Then finish filling in the table by simple subtraction from the totals for each row.
4) Calculate
5.) Calculate d.f. (4 - 1)(2 - 1) = 3
6) Significance level at .05
7) X^2 value in table = 7.81
8) Compare the computed X^2 and the X^2 table value.
The computed value is larger, therefore the H[0] is rejected. The relationship between milkweed species A and B is statistically significant.
EXAMPLE 2: Is there a position effect in the cage?
1) State the Null Hypothesis: There is no difference in preference of the plant placed on the left side or right side of the cage for egg laying.
2) Calculate ECF, and 3.) Fill the Expected Data Table
Calculate the ECF for Left Side - Trial 1 = (396 x 148) / 594
ECF = 99
Calculate the rest of the ECF s for the Left Side. Then finish filling in the table by simple subtraction from the totals for each row.
4) Calculate
5) Calculate d.f. (4 - 1)(2 - 1) = 3
6) Significance level at .05
7) X^2 value in table = 7.81
8) Compare the computed X^2 and the X^2 table value.
The computed value is smaller, therefore failed to reject the H[0]. The there is no position effect of the plant placed on the left or the right side of the cage.
Chi Square Distribution Table
M. C. Fleming and J. G. Nellis, Principles of Applied Statistics, New York: Routledge, 1994, p. 400.
A. D. Rickmers and H. N. Todd, Statistics: An Introduction, St. Louis: McGraw-Hill, 1967, p. 585. | {"url":"http://monarchwatch.org/read/articles/eggs.htm","timestamp":"2014-04-16T21:56:01Z","content_type":null,"content_length":"27159","record_id":"<urn:uuid:14f325bb-aa23-47e3-b1a3-acfeeede4ba2>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00389-ip-10-147-4-33.ec2.internal.warc.gz"} |
Secure Quantum Network Proved Robust by 21 Months of Use - Industry News
Secure Quantum Network Proved Robust by 21 Months of Use
Science & Technology
Posted: December 2, 2011 01:51PM
Secure communication with modern computers relies on mathematics so intense, it is effectively impossible for a hacker to break it without the key. How the encryption system works is to multiply the
data by a large prime number. There is no known way to predict prime numbers, so cracking the encryption would require testing every smaller number. For modern computers this would take more time
than the age of the Universe, but for quantum computers, this is not the case. Already an algorithm exists to efficiently crack encryption with a quantum computer, so how would a quantum network
transmit information securely?
Quantum key distribution (QKD) is one possible method and takes advantage of observation destroying quantum states. The states, in this case, are the orientations of a photon. For now, you can
picture the photon as a compass; the needle is either vertical (↑) or horizontal (→), or at a diagonal (↗ or ↘). If Alice and Bob (told you they’d get back together) want to transmit information
to each other, Alice, starts by randomly sending a bit to Bob, with one of those orientations. Now Bob has to decide what orientations he is measuring, because ↑ and → are only recorded correctly if
he measures rectilinearly, and the diagonal bits (↗ or ↘) would be randomly 0s or 1s, and similarly ↑ and → would be random when Bob measures on a diagonal. Alice continues to send random bits
like this, changing the orientation and the reference for what is 0 and 1 each time.
With all the random bits sent and measured, Bob tells Alice, without any quantum mechanics involved, what he measured each time; a 0 or 1. Alice replies with if he is right or wrong for each bit, and
Bob, knowing how he measured it each bit, then knows how each bit was sent. (The only way he would be right is if he measured the same way she sent it.) They both now throw out whenever he was
incorrect, and what’s left makes the shared key between them, for secure communication.
This method is secure because every time a photon is measured, the photon is disturbed in such a way that either Alice or Bob can notice it. Any would-be eavesdroppers are immediately discovered.
Now researchers know this method can work in real life. From March 2009 to January 2011, a QKD network was used by Swiss scientists at ID Quantique and the scientists at CERN. The only times the
network was disrupted, an external event like a power outage caused it, and not a fault of the network or QKD system itself. | {"url":"http://www.overclockersclub.com/news/30091/","timestamp":"2014-04-16T22:17:15Z","content_type":null,"content_length":"23201","record_id":"<urn:uuid:06403e77-d5bb-4e39-858e-6c1af285f506>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00414-ip-10-147-4-33.ec2.internal.warc.gz"} |
There are many cases where you would like to be able to click on and select an entity in the 3D world. In order to do this we need to take the screen position of the mouse pointer and convert it to a
position in the world and a direction. This gives us a ray that we can then use to check for collisions with the entities in our world. This ray travels from the users eye through the screen (at the
mouse position) and into the 3D world.
When we render an entity it is transformed from model space into world space via the world matrix set for each entity (see Matrices). It is then transformed into camera space using the view matrix
before finally being rendered with perspective to the 2D screen using the projection matrix. When we click on the 2D screen with the mouse we need to do the reverse, we need to take the 2D point and
convert it into a position and ray in our 3D world.
Creating the ray
We need to take the mouse position and convert it into a ray. The mouse position returned by the Windows API has co-ordinates ranging from top left of the screen (0,0) to bottom right (width,
height). We need to adjust this into a position in our world. So firstly we take into account the mouse range and use the projection matrix values:
D3DXVECTOR3 v;
v.x = ( ( ( 2.0f * sx ) / w ) - 1 ) / matProj._11;
v.y = -( ( ( 2.0f * sy ) / h ) - 1 ) / matProj._22;
v.z = 1.0f;
Where sx and sy are the mouse screen positions. w is the width of the screen and h the height. matProj is the original projection matrix you set (if you have not stored matProj you can retrieve it by
using the device->GetTransform(..) method).
Note: it is important that the width and height values above are correct. You need to use the size of the back buffer which may not be the same as the window size (due to menu bars, borders etc.).
What I normally do is after creating the device I call device->GetViewport( &m_mainViewport ) and store it. The viewport structure holds the correct back buffer width and height.
The next transform determining the 2D representation is the view matrix, so we need to apply the inverse of this matrix to create our ray:
D3DXMATRIX m;
D3DXVECTOR3 rayOrigin,rayDir;
D3DXMatrixInverse( &m, NULL, &matView );
// Transform the screen space pick ray into 3D space
rayDir.x = v.x*m._11 + v.y*m._21 + v.z*m._31;
rayDir.y = v.x*m._12 + v.y*m._22 + v.z*m._32;
rayDir.z = v.x*m._13 + v.y*m._23 + v.z*m._33;
rayOrigin.x = m._41;
rayOrigin.y = m._42;
rayOrigin.z = m._43;
We have now created a ray in world space from our mouse 2D position. It has a position in the world (rayOrigin) and a direction (rayDir). The direction is a vector defining the direction from the eye
through the screen into the 3D world. rayOrigin is at the position of the camera.
The next step is to collide this ray with the entities in our world. Normally you would do a bounding box and / or bounding sphere test first to trivially reject entities before doing an actual ray
versus mesh intersection test.
Ray - Mesh Intersection test
Direct3D provides a useful function called D3DXIntersect that takes a mesh and a ray and determines if the ray has hit the mesh. In addition it can return the distance to the collision and other
useful values.
The most important thing to remember is that you must do the intersect calculations in model space. So for each world entity we want to test against we must convert the ray into that entities
graphics model space. To do this we take the inverse of the entities world matrix and apply it to our ray:
// Use inverse of matrix
D3DXMATRIX matInverse;
// Transform ray origin and direction by inv matrix
D3DXVECTOR3 rayObjOrigin,rayObjDirection;
We can now call the intersect function on our untransformed graphic mesh data:
BOOL hasHit;
float distanceToCollision;
D3DXIntersect(m_mesh, &rayObjOrigin, &rayObjDirection, &hasHit, NULL, NULL, NULL, &distanceToCollision, NULL, NULL);
• hasHit is true if a collision has occurred.
• distanceToCollision gives the distance from the ray origin to the collision point.
The other parameters allow you to get more detailed values determining the actual collision point.
If you do not have a Direct3D mesh you can use D3DXIntersectTri which takes an array of vertex positions and uses those to calculate collision.
In your world model you will loop through all entities in your world doing the above ray intersection test. If a collision occurs remember the distance and keep looping. If you find another collision
closer then that is the one to use (you could click on an entity with another behind it but really just want to pick the closest one).
There are many ways to optimise picking. Firstly there is no need to test against entities outside of the viewing frustum. Secondly you should test against an entities AABB first and trivially reject
those that do not collide - this is much quicker (you could also check against a bounding sphere or some other bounding volume). With very complex geometry you could even do your picking against a
lower density mesh that is never rendered but purely used for collision purposes.
Further Reading | {"url":"http://www.toymaker.info/Games/html/picking.html","timestamp":"2014-04-20T23:28:04Z","content_type":null,"content_length":"41112","record_id":"<urn:uuid:93eb8df4-34bb-42ab-bd4c-8ac8478115b2>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00645-ip-10-147-4-33.ec2.internal.warc.gz"} |
An LC Circuit Consists Of A Capacitor, C = 1.68 ... | Chegg.com
An LC circuit consists of a capacitor, C = 1.68 µF, and an inductor, L = 2.68 mH. The capacitor is fully charged using a battery and then connected to the inductor. An oscilloscope is used to measure
the frequency of the oscillations in the circuit. Next, the circuit is opened, and a resistor, R, is inserted in series with the inductor and the capacitor. The capacitor is again fully charged using
the same battery and then connected to the circuit. The angular frequency of the damped oscillations in the RLC circuit is found to be 29.5% less than the angular frequency of the oscillations in the
LC circuit.
a) Determine the resistance of the resistor.
b) How long after the capacitor is reconnected in the circuit will the amplitude of the damped current through the circuit be 55.5% of the initial amplitude?
c) How many complete damped oscillations will have occurred in that time? | {"url":"http://www.chegg.com/homework-help/questions-and-answers/lc-circuit-consists-capacitor-c-168-f-inductor-l-268-mh-capacitor-fully-charged-using-batt-q2431495","timestamp":"2014-04-18T15:01:40Z","content_type":null,"content_length":"19489","record_id":"<urn:uuid:b5bbca16-eb9c-4cf4-873a-993a6c08a244>","cc-path":"CC-MAIN-2014-15/segments/1397609533689.29/warc/CC-MAIN-20140416005213-00064-ip-10-147-4-33.ec2.internal.warc.gz"} |
URM v. Non-URM Empirical Comparison
Here is a statistical analysis that I conducted using self-reported figures as ascertained from
. I do not wish to present any conclusive evidence or make any judgments regarding the use of Affirmative Action in law school admissions, I simply wish to provide some numbers that may assist some
in making a determination about the issue.
I took the LSAT and GPA scores of the first 9 students who had been accepted to Yale c/o '09, controlling only for under-represented minority status (i.e. URM and non-URM indentification). I consider
these scores to be randomized because there is not any order to how the pseudonyms are listed on the website, so I did not deem it necessary to further randomize the pool. Then I calculated the mean
scores for each group and computed standard deviation for each group as well. I will not list the names of the students I used from the website in order to protect their anonymity.
URM LSAT mean= 170.7
URM LSAT standard deviation= 4.658
URM GPA mean= 3.895
Non-URM LSAT mean= 175.2
Non-URM LSAT standard deviation= 3.527
Non-URM GPA mean= 3.883
Note: I realize that there is a flaw in the methodology of this statistical analysis in that self-reported scores are subject to inflation or deflation. Therefore, I do not offer this as conclusive
evidence of any kind of bias in admissions but the number trend is statistically significant given these limitations.
Feel free to comment or provide any additional insight pertaining to this matter. Thank you. | {"url":"http://www.lawschooldiscussion.org/index.php?topic=68507.msg1613468","timestamp":"2014-04-20T22:33:58Z","content_type":null,"content_length":"36040","record_id":"<urn:uuid:81de3c34-57e1-4149-b0ee-326e36cbc4ad>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00566-ip-10-147-4-33.ec2.internal.warc.gz"} |
[FOM] Inconsistent Systems
Harvey Friedman hmflogic at gmail.com
Fri Sep 13 22:08:27 EDT 2013
On Fri, Sep 13, 2013 at 4:56 PM, Alan Weir <Alan.Weir at glasgow.ac.uk> wrote:
> Regarding Harvey Friedman's sketch of the triviality of a form of naive
> set theory- FOM Digest, Vol 129, Issue 17- it is well known that Curry's
> paradox trivialises naive set theory even in quite weak relevant logics
> without any appeal to the logic of negation. Those who seek to maintain
> that naive set theory is nonetheless still consistent or (this is the more
> usual line) at least non-trivial, often see the culprit as that principle
> of contraction which in rule form (sequent form) natural deduction goes:
I had a feeling that what I sketched in
http://www.cs.nyu.edu/pipermail/fom/2013-September/017596.html was known,
and I am glad I didn't make that one of my numbered postings! Thanks also
to Avron, Arthan, and Marcos for pointing this out.
Of course producing a non-classical framework in which naive set theory is
> consistent (or at least non-trivial) is one thing, and certainly can be
> done. Producing a framework which is not so restricted that it cripples
> standard mathematics is another, much more challenging, thing.
I believe that "it cripples standard mathematics", and even "profoundly
cripples standard mathematics". But I would be fascinated if a good case
can be made that it does NOT cripple standard mathematics, and the
development of a suitable treatise in which mathematics is actually
developed with such "logics".
Of course, it would also be an interesting challenge to prove that such a
suitable treatise cannot exist.
So I will operate on the working assumption that CA(no) and other forms of
CA simply cannot be used for the foundations of mathematics.
Now I turn to a related matter which I also hesitate to put in my numbered
I think I need to clarify a point here, that I didn't make with regard to
CA(no). Obviously, subsystems of CA(no), or even CA, can be used for f.o.m.
So what am I talking about?
What I mean is, can we use the entire set of axioms, but with a weakened
logic - where the weakened logic is sufficient to do ordinary mathematics?
It appears that the answer is no for CA(no).
Now, let's look at ZFC. Clearly we can complain about the use of classical
logic to derive its hypothetical inconsistency. And there is a perfectly
good intuitionistic ZF - see The Consistency of Classical Set Theory
Relative to a Set Theory with Intuitionistic Logic, J. of Symbolic Logic,
Vol. 38, No. 2, (1973), pp. 315-319. I think I called this IZF (I forget).
THEOREM?? (finitary). Suppose ZFC is inconsistent. Then IZF without
absurdity and without negation proves every formula without absurdity and
without negation.
I leave this to the experts to complete. And perhaps discuss also whether
anything can be salvaged from an inconsistency in ZFC other than the
obvious move of going to subsystems in the usual sense.
> (I note also that Prof. Friedman's system CA(no) might perhaps be more
> naturally called a theory of naive properties, since there is no
> extensionality axiom. Hartry Field has given a consistency proof for such a
> system in quite a strong non-classical logic.)
> Did Field or others see just what happens if we try to use this "strong
non-classical logic" for actual mathematics?
Harvey Friedman
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Complex no again
December 11th 2012, 05:10 AM
Complex no again
I really need help here!
Using the triple angle identities for sin, cos and tan, Solve the equation
1.) 4x^3 - 3x = -(1/sqrt2)
At first instinct I saw that it resembled the triple angle identity for cosine so I had cos3x=-(1/sqrt2), arccos(1/sqrt2) gave me pi/4 and then cosx is pi/12. However, the answer had cos(11pi/12)
and cos(5pi/12), how did they get that?
2.) Solve x^3 - 3sqrt(3)x^2 -3x + sqrt(3) = 0. I couldn't see how it represented any of the triple identities. I thought tan3x is possible but I have problems with the coefficients, they don't
Please help me, thank you so very much!
December 11th 2012, 06:34 AM
Re: Complex no again
When you are solving an equation like cos(3t) = - 1 / sqrt(2), you should first look at the unit circle and determine the sign and the approximate values of 3t. Use arccos only in the end to get
the numerical value, and only if the the argument of cosine is not a multiple of 30 or 45 degrees.
In this case, if you draw a vertical line through x = - 1 / sqrt(2), it intersects the unit circle at 3pi/4 and 5pi/4. So,
$3t = 3\pi/4+2\pi k$ or
$3t = 5\pi/4+2\pi k$
where $k\in\mathbb{Z}$. This means
$t = \pi/4 + 2\pi k/3$ or
$t = 5\pi/12 + 2\pi k/3$
Now find all t from 0 to 2pi. Some of them will generate the same solution x = cos(t) to the original equation.
December 11th 2012, 06:41 AM
Re: Complex no again
Interesting way to do algebra problems. New to me.
As for the second, I think it does tally up with the triple tangent formula. Looks that way to me. Hope this helps. | {"url":"http://mathhelpforum.com/number-theory/209584-complex-no-again-print.html","timestamp":"2014-04-17T04:09:37Z","content_type":null,"content_length":"6025","record_id":"<urn:uuid:d7fd68f2-0d1a-4f5d-8d30-7c073e5e355f>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00236-ip-10-147-4-33.ec2.internal.warc.gz"} |
Taylor’s Expansion for Composite Functions
The Scientific World Journal
Volume 2013 (2013), Article ID 536280, 7 pages
Research Article
Taylor’s Expansion for Composite Functions
^1Nhatrang Educational College, 01 Nguyen Chanh Street, Nhatrang City, Vietnam
^2Department of Mathematics, University of Architecture of HoChiMinh City, 196 Pasteur Street, District 3, HoChiMinh City, Vietnam
Received 4 August 2013; Accepted 28 August 2013
Academic Editors: A. Agouzal and J.-S. Chen
Copyright © 2013 Le Thi Phuong Ngoc and Nguyen Anh Triet. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
We build a Taylor’s expansion for composite functions. Some applications are introduced, where the proposed technique allows the authors to obtain an asymptotic expansion of high order in many small
parameters of solutions.
1. Introduction
Let and , .
Let two functions and, which depend on and , be defined as follows: where where notation stands for the norm in .
The function has the form of a composite function as follows: where where notation stands for the scalar product in .
Under the above assumptions, in order to obtain an asymptotic expansion of high order in many small parameters of solutions in recent papers [1–4], we need to solve the following problem.
Problem 1. Establish the functions (independent of ) such that for being small enough, it means that where the constant is independent from .
By Taylor’s expansion for , it implies that
However, for which is given in (3), it is very difficult to calculate , and so, the Problem 1 cannot proceed. Whichmethod can be used for solving the Problem 1? To answer, let us note that the
function has the form of a composite function;this viewsuggests thatwe need to construct the Taylor-Maclaurin expansion of the composite function. So, first in Section 2 we solve the following
Problem 2. Let be an open subset of and . Letand.Seek the representation formula for , such that for being small enough, where are calculated from the values of the given functions and of their
derivatives at a suitable point.
Next in Section 3, as an application of the method used, we study the Problem 1. This technique is also a great help for the authors to obtain an asymptotic expansion as they want in recent papers [1
2. Solving the Problem 2
We use the following notations. For a multi-index and , we put
The following lemma is useful to solve the Problems 1 and 2.
Lemma 3. Let and . Then where the coefficients depending on are defined by the recurrent formulas
The proof of Lemma 3 can be found in [2].
Now, using Taylor’s expansion of the function around the point , we obtain that
Similarly, we use Maclaurin’s expansion of as follows:
Substituting (13) into (12), we get where
Applying Lemma 3, with , it implies that
Clearly, the Problem 2 is solved with
3. Solving the Problem 1
As an application of the method used in Section 2 for , and , for eachfixed , , for eachfixed , for eachfixed , we now investigate the Problem 1.
For eachfixed , using Taylor’s expansion of the function around the point up to order we obtain that where is small enough, .
Next, the precise structure of the representation formulas for will be needed below to continue.
For each fixed we have the following.
(i) The representation formula for .
We rewrite as follows in which
It is similar to ; we write where
For eachfixed we have the following.
(ii) The representation formula for .
Similarly, in which where
Then(22), (24), (26), (28), and (30) imply that
We also need the following lemma.
Lemma 4. For all , then where
Proof of Lemma 4. We have
Note that , so
Combining (35) and (36) yields where
Lemma 4 is proved.
Substituting (37) into (20), we obtain where
Since , hence . Then from (39) and (40)[1], we get for is small enough; consequently the Problem 1 is solved.
Remark 5. (i) This result improves the one in [1], for .
(ii) In case of , where , this result is also obtained in [3].
The authors wish to express their sincere thanks to the referees for their valuable comments and important remarks. This research is funded by Vietnam National University HoChiMinh City (VNU-HCM)
under Grant no. B2013-18-05. | {"url":"http://www.hindawi.com/journals/tswj/2013/536280/","timestamp":"2014-04-19T19:03:03Z","content_type":null,"content_length":"725151","record_id":"<urn:uuid:81dbfa46-c96e-4a15-ade0-6ee3576ffab7>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00392-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Grey Labyrinth
Three Hats
At first glance, this problem appears to be impossible to solve. Contributing to this is the feeling that the King's only real clue- that there is at least one blue hat- is useless since the sage can
clearly see that there are at least two blue hats.
Don't feel bad if you sat stuck on this one for a while: as the puzzle clearly states, so did the three wisest sages in the kingdom. It is this fact that allowed our sage to give his answer. In
truth, any one of them would have come up with it, given enough time. Why?
Consider a situation which we knew was not the case- that there was exactly one blue hat. What would happen? There would be a split second of pondering by the person wearing that hat, and he would
say "I am wearing a blue hat." No real puzzle there, but of course there wasn't just one blue hat. The important fact is that everyone knew there was not one blue hat. But more importantly than that,
everyone knew, or could quickly figure out that everyone else knew this (by the fact that answer was did not come out in the first few seconds.)
This leaves everyone wondering, "Are there two or three blue hats?"
Consider this less obvious situation- that there were exactly two blue hats. This seems a very real possibility at first, after all, we can see exactly two blue hats. So everyone sits and thinks- for
a little while. But if there are only two hats, then two people see one blue and one white hat. These two people will very quickly, by virtue of the other's silence, rule out the possibility that
there is only one blue hat. One of these two lucky sages would cry blue within a few short minutes, if that long.
There is only one case which forces the three sages to sit in silence- three blue hats. Our sage, through his sharp wits was the first to reach this conclusion. | {"url":"http://greylabyrinth.com/solution/puzzle007","timestamp":"2014-04-20T20:55:54Z","content_type":null,"content_length":"7174","record_id":"<urn:uuid:52b5d1cd-20a4-44e1-823f-7cb6df2295b9>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00368-ip-10-147-4-33.ec2.internal.warc.gz"} |
Summary: POLYNOMIAL AND RATIONAL MATRIX INTERPOLATION:
Panos J. Antsaklis
Dept ofElectrical Engineering
University ofNotre Dame
Notre Dame, IN 46556
Email: antsaklisatn.ece.nd.edu
Abstract: In this paper, a theory of polynomial and rational
matrix interpolation is hintrodued and applied to problems in
systems and control. The polynomial matrix interpolation
theory is first outlined and then applied to solving matrix
equations; it is also used in pole assignment and other control
problems. Rational matrix interpolation is also discussed and
it is used to solve rational matrix equations including the model
matching problem.
A theory of polynomial and rational matrix interpolation
is briefly outlined in this paper and its application to certain
systems and control problems is discussed; full details can be
found in [21]. | {"url":"http://www.osti.gov/eprints/topicpages/documents/record/607/1764724.html","timestamp":"2014-04-21T10:07:29Z","content_type":null,"content_length":"8002","record_id":"<urn:uuid:041b0ef5-8157-4071-948c-a0e9f63cfb22>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00341-ip-10-147-4-33.ec2.internal.warc.gz"} |
Solve Real-World Problems by Using Strategies and a Plan
1.16: Solve Real-World Problems by Using Strategies and a Plan
Practice Applications Using Linear Models
Have you ever gone on a plane trip? Take a look at this dilemma involving flights, times and friends.
Kevin and Laila sat talking at lunch. Then Carmen came over.
“What are you guys talking about?” Carmen asked.
“We were talking about what we did this summer. I went to Yellowstone camping and Kevin did this really great project with a group of kids making a garden. What about you? What did you do with your
summer?” Laila asked Carmen as she took a drink of milk.
“I went to see my Grandparents. It was a great time, but I barely made it on the day of my flight,” Carmen said munching a carrot.
“What happened?” Kevin asked.
“Well, it started out fine. I had a 9 pm flight. I knew that I had to be at the airport 2 hours before the flight and that we live one hour from the airport. I needed $1 \frac{1}{2}$
Kevin looked at Laila.
“You should have had plenty of time,” Kevin said.
How does Kevin know this? Can you follow Kevin’s thinking? Why does Kevin make this statement? To figure this out, you will need to apply your problem solving skills.
When problem-solving, you will use strategies as part of a plan. For each situation, you will be asked to read and understand a given problem. You will make a plan to solve by choosing an appropriate
strategies. There are multiple ways to solve a word problem. Therefore, you will need to consider and compare different approaches for each problem given.
First, you read the problem.
When you read to understand a problem, you are working to determine what the problem is asking you to do. It helps to highlight the question in the problem. You may want to also underline clue words
that may help you with planning and strategy.
A lizard ate five hundred flies on five consecutive nights. Each night he ate twenty-five more than the night before. How many flies did the lizard eat each night?
For this problem, you are to determine the number of flies the lizard ate each night. Here is what you are told:
• You are told that the lizard ate a total of five hundred flies over the course of five nights.
• You are told that the lizard eats twenty-five more flies each night than the night before.
• You should know the word consecutive means a logical sequence or succession. In this case, it means one night after the other.
Make plan to solve the problem.
Use a verbal model to write and solve an equation to determine the unknown number of flies eaten each day.
You are told that the lizard ate a total of five hundred flies in five days. You are also told that each night he eats twenty-five more than the night before. To determine the number of flies
consumed each night, you must first determine the number of flies the lizard ate the first night. After determining the number of flies consumed the first night, add twenty-five more each day to get
the daily total.
Verbal Model:
number of flies eaten on night one + (number of flies eaten on night one + twenty-five) + (number of flies eaten on night one + twenty-five + twenty-five) + (number of flies eaten on night one +
twenty-five + twenty-five + twenty-five) + (number of flies eaten on night one + twenty-five + twenty-five + twenty-five + twenty-five) = total number of flies eaten over five nights (500)
Let “ $x$
The next step is to solve the problem.
$x + (x + 25) + (x + 25 + 25) + (x + 25 + 25 + 25) + (x + 25 + 25 + 25 + 25) = 500$
$x + (x + 25) + (x + 25 + 25) + (x + 25 + 25 + 25) + (x + 25 + 25 + 25 + 25) &= 500\\x + (x + 25) + (x + 50) + (x + 75) + (x + 100) &= 500\\5x + 250 &= 500$
Next, we solve the equation. Subtract 250 from 500 and divide by 5.
The lizard consumed 50 flies the first night. To determine the number of flies the lizard ate on nights two, three, four, and five, substitute 50 for “ $x$
$x + (x + 25) + (x + 50) + (x + 75) + (x + 100) &= 500\\50 + (50 + 25) + (50 + 50) + (50 + 75) + (50 + 100) &= 500\\50 + 75 + 100 + 125 + 150 &= 500$
You can see that on night one, the lizard ate 50 flies.
On night two, the lizard consumed 75 flies.
On night three, the lizard ate 100 flies.
On night four, the lizard ate 125 flies.
On the last night, the lizard ate 150 flies.
Check the Results
You can check your work, by adding the number of flies consumed each night. The sum should be equal to five hundred.
$50 + 75 + 100 + 125 + 150 = 500$
Night One: 50
Night Two: 75
Night Three: 100
Night Four: 125
Night Five: 150
Now look at this situation and answer each question.
A ten year olds’ heart beats approximately 85 times per minute. How many times does the heart beat in 24 seconds?
Example A
What is the problem asking you to do?
Solution: Knowing that the heart beats 85 times per minute, determine the number of heart beats in 24 seconds.
Example B
What is the plan?
Solution: Since you are being asked to determine an unknown rate, write a proportion to solve.
Example C
Can you write an equation and solve this problem?
Solution: Let “ $x$
$& \ \underline{\;\; 85 \ beats\;\;} = \underline{\;\;\;\;\;\;\; x \;\;\;\;\;\;\;}\\& \ 60 \ seconds \quad \ 24 \ seconds\\& \qquad 85(24) = 60(x)\\& \ \quad \underline{\;\; 2,040 \;\;} = \underline
{\;\; 60x \;\;}\\& \ \qquad \ 60 \qquad \quad \ 60\\& \qquad \qquad \ x=34 \ beats$
Now let's go back to the dilemma from the beginning of the Concept.
First, why did Kevin make that statement?
Kevin figure out that Carmen should have left for the airport at 6 pm. If she only needed $1 \frac{1}{2}$
Here is the breakdown of her time.
9 pm flight – 2 hours check in = 7 pm
7 pm – 1 hour drive time = 6 pm
$6 - 1 \frac{1}{2}$
Carmen should have left her home at 4:30 pm to be on time for the flight.
a comparison between two quantities. Ratios can be written in fraction form, with a colon or by using the word “to”.
formed when two ratios are equivalent. We compare two ratios, they are equal and so they form a proportion.
Guided Practice
Here is one for you to try on your own. Notice that some key points are underlined for you.
A train’s caboose is 12 feet long . Each of the train’s eight cars are twice the length of the caboose . Determine the length of the entire train.
Ask: What is the problem asking you to find out?
You are to determine the entire length of the train. You were told some information.
• The caboose is 12 feet long.
• There area of an additional eight cars.
• Each car is twice the length of the caboose.
Make a plan to solve the problem.
You can draw a diagram and use a verbal model to visualize the information given in the problem. Then, write an equation to determine the length of the entire train.
Verbal Model:
Eight trains twice the length of the caboose + the length of the caboose = entire length of the entire train
Let “ $x$
$8(2 \cdot 12) + 12 = x$
$8(2 \cdot 12) + 12 &= x\\8(24) + 12 &= x\\192 + 12 &= x\\204 &= x$
The entire train is 204 feet.
Video Review
Khan Academy Problem Solving Strategies
Directions: Read each problem and then solve each problem.
Ted has a collection of rare coins. He already had 34 coins in his collection. The first week, Ted purchases 1 new coin. During the second week, Ted purchases 4 coins. During the third week, Ted adds
9 new coins to his collection. At this rate how many weeks will it take Ted to collect 125 coins?
1. Which strategy should Ted use to solve this problem?
2. What could Ted draw to help him with his solution?
3. How long will it take Ted to collect 125 coins?
Savannah wants to buy a pair of jeans that cost $59.00. They are on sale for 25% off.
4. Which strategy could Savannah use to calculate the price?
5. What is the amount of the discount?
6. What is the sale price?
Carlos was in charge of organizing cookies for a bake sale. He organized them into bundles of six cookies. When he was done, he had 15 bundles of cookies. How many cookies did he start with?
7. Which strategy could you use to solve this problem?
8. Write an equation to describe the problem.
9. Solve the equation.
10. How many cookies did Carlos begin with?
Veronica made brownies. She made twice as many brownies as Carlos had cookies. How many brownies did she make?
11. Which strategy could you use to solve this problem?
12. Write an equation to describe the problem.
13. Solve the equation.
14. How many brownies did Veronica make?
If Veronica sold half of the cookies that she made, how many would be sold? If she charged $1.50 per brownie, how much money would she make?
15. Which strategy could you use to solve this problem?
16. How much money would she make?
Files can only be attached to the latest version of Modality | {"url":"http://www.ck12.org/book/CK-12-Middle-School-Math-Concepts-Grade-8/r12/section/1.16/","timestamp":"2014-04-18T03:15:47Z","content_type":null,"content_length":"152666","record_id":"<urn:uuid:3b97de62-e365-487f-806b-fae058a2af4c>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00175-ip-10-147-4-33.ec2.internal.warc.gz"} |
Characterization of Measureable Sets
up vote 0 down vote favorite
Every countable union of rectangles in R2 is a Lebesgue measurable set. Is the converse true, too?
Specifically, I wonder whether the following statement is true:
Let A be a set in the unit square that is Lebesgue measurable. Then there a countable collection of rectangles and a null set such that A is equal to the union of the rectangles and the null set.
fa.functional-analysis measure-theory
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1 Answer
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No, see here for a counter-example (it's a variant of the Cantor set which has non null Lebesgue measure and does not contain any interval).
up vote 0 down vote And if you want a counter-example for $\mathbb{R}^2$ instead of $\mathbb{R}$, just cross it with an interval.
Yes, that should help answer it. The argument will then be that the symmetric difference of any nontrivial interval with the SVC set is not a null set. – Stephan Dec 7
'11 at 0:41
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Not the answer you're looking for? Browse other questions tagged fa.functional-analysis measure-theory or ask your own question. | {"url":"http://mathoverflow.net/questions/82836/characterization-of-measureable-sets","timestamp":"2014-04-19T10:13:21Z","content_type":null,"content_length":"45780","record_id":"<urn:uuid:8212bd61-7550-4a59-b54d-22f3dcb15263>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00503-ip-10-147-4-33.ec2.internal.warc.gz"} |
FOM: Test Case for "Elementary" Proofs
Lou van den Dries vddries at math.uiuc.edu
Thu Dec 11 19:16:59 EST 1997
There is a very elementary proof of the n=1 case of Dirichlet's theorem:
given k > 1 there are infinitely many primes in the progression
[k+1, 2k+1, 3k+1, ...]. I can't resist indicating it (as i found it
myself in highschool, though of course it was well known as I learned
later from Dixon's history). The case k=10 will give the main idea:
if p is a prime number dividing x^{10}-1 but not x^5 -1 and also not
x^2 -1 (where x is some integer > 1), then by Fermat's little theorem
10 must divide p-1, that is, p must occur in the arithmetic progression
11, 21, 31, 41, ......
Now, for x > 1 the prime factors of x^{10} -1 that are not among the
prime factors of x^5 -1 or x^2 -1 are prime factors of
(x^{10} -1)/ (x^5 -1)(x^2 -1) = x^3 + ..., -1.
By an euclidean type argument (as in "there are arbitrarily many primes")
one shows very easily that by letting x run through the integers > 1
one can produce given arbitrarily large primes dividing the
right hand side in the display above (by choosing x > 1 suitably).
These primes will indeed be in the above arithmetic progression.
(There are a few things to check here, and in general the argument
needs a careful counting of the degree of a polynomial, which turns out
to be the cyclotomic polynomial whose roots are the primitive k^th
roots of unity, but this is no real problem, and certainly elementary
according to any standards, I think.)
Actually, I already made a mistake, I now see: since x-1 is a common
factor of x^5 -1 and x^2 -1, I should compensate for this by
multiplying the LHS above by (x-1) (fortunately, since otherwise
the division would leave a remainder). So replace display above by
(x^{10} -1) (x-1)/ (x^5 -1)(x^2 -1) = x^4 + ... + 1
Sorry. (In high school I didn't know about cyclotomic poly's, but I
did know that the division above would work for all x, and would
produce a number that grows like x^{\phi(k)}, where \phi(k) is
Euler's phi function. And that's enough.)
Apologies for the nonfom nature of the message. -Lou van den Dries-
More information about the FOM mailing list | {"url":"http://www.cs.nyu.edu/pipermail/fom/1997-December/000478.html","timestamp":"2014-04-21T08:32:18Z","content_type":null,"content_length":"4386","record_id":"<urn:uuid:26f24294-502d-48ee-8180-09a0e2c90068>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00159-ip-10-147-4-33.ec2.internal.warc.gz"} |
Derivative of Geometric Progression
From ProofWiki
Let $x \in \R: \left|{x}\right| < 1$.
$\displaystyle \sum_{n \mathop \ge 1} n x^{n-1} = \frac 1 {\left({1-x}\right)^2}$
$\displaystyle \sum_{n \mathop \ge 1} n \left({n+1}\right) x^{n-1} = \frac 2 {\left({1-x}\right)^3}$
We have from Power Rule for Derivatives that:
$\displaystyle \frac {\mathrm d} {\mathrm d x} \sum_{n \mathop \ge 1} x^n = \sum_{n \mathop \ge 1} n x^{n-1}$
But from Sum of Infinite Geometric Progression:
$\displaystyle \sum_{n \mathop \ge 1} x^n = \frac 1 {1-x}$
The result follows by Power Rule for Derivatives and the Chain Rule applied to $\dfrac 1 {1-x}$.
Why should the more complicated second derivative result be a corollary of the (on the surface) easier first derivative result?
Because the clever bit is thinking of the differentiation technique in the first place. Applying it to the second result is just an exercise in application. | {"url":"http://www.proofwiki.org/wiki/Derivative_of_Geometric_Progression","timestamp":"2014-04-20T00:38:28Z","content_type":null,"content_length":"25541","record_id":"<urn:uuid:ad5f9e4d-fdd1-4405-b28f-a7ac4d4eb463>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00587-ip-10-147-4-33.ec2.internal.warc.gz"} |
Which year was Daisy Lowe born in?
You asked:
Which year was Daisy Lowe born in?
Say hello to Evi
Evi is our best selling mobile app that can answer questions about local knowledge, weather, books, music, films, people and places, recipe ideas, shopping and much more. Over the next few months we
will be adding all of Evi's power to this site.
Until then, to experience all of the power of Evi you can download Evi for free on iOS, Android and Kindle Fire. | {"url":"http://www.evi.com/q/which_year_was_daisy_lowe_born_in","timestamp":"2014-04-19T08:02:39Z","content_type":null,"content_length":"52128","record_id":"<urn:uuid:affeb0bb-62c6-4d9e-ad11-2ee08bb4f122>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00622-ip-10-147-4-33.ec2.internal.warc.gz"} |
Capital Asset Pricing Model
The capital asset pricing model functions with the understanding that investors need additional compensation when they invest in riskier ventures. It also takes into consideration the time value of
money. The formula for a CAPM calculator uses three pieces of information to calculate the necessary return; the risk free rate, the beta of the investment and the expected market return. The beta of
a stock is the deviation that it has compared to the market. The return of the investment should be the risk free rate plus the result of the beta times the difference between the expected market
return and the risk free rate.
Many financial planners use the capital asset's pricing model to decide whether the additional risk of a stock is worth adding to a client's portfolio. Sometimes the CAPM is used to judge the total
portfolios risk and decide if changes need to the portfolio are necessary.
As you apply the CAPM model, consider the following:
1. Know the capital asset pricing model definition so you can determine other capital investments besides stock.
2. Use the capital asset pricing model formula to understand how diversification of asset's risk classification reduces risk.
3. of the major advantages of CAPM is the fact that it provides a formula where financial decisions are based on a risk to reward ratio.
Look for capital asset pricing model calculator software for portfolio risk management
You can purchase software for CAPM calculation. The type of software you purchase varies with your need. Investment portfolio management, stock and option evaluation and business expenditures all
come on different programs. Occasionally a program provides both portfolio management and individual investment analysis.
Get information that integrates the CAPM analysis with the modern portfolio theory
Many programs offer CAPM analysis but fail to do a practical integration of the asset into the portfolio. The more diversified the portfolio, the greater the reduction of the systematic risk.
Unfortunately, diversification isn't enough so the capital asset pricing model formula filled in the blanks and finalized the theory. You can learn the use of the model to improve your business savvy
and aid the client is choosing the right stock.
An Introduction to Investment Theory
by William N. Goetzman is available online for free. It covers not only CAPM but other investment theory. Brokers benefit from the book Modern Investment Theory, which comes with software and tests
on the investment theory and asset management.
Professionalize your brokerage business using CAPM analysis
Today's broker needs to be more innovative and know leg able to capture high dollar clients. The use of software reduces the workload and increases the attractiveness of the presentation. These
factors allow you to handle more clients with ease and impress those that you already have.
• All financial planners and brokers should purchase their own Capital Asset Pricing Model (CAPM) software for use with clients. Free downloads are intended for an individual's personal use. | {"url":"http://www.business.com/finance/capital-asset-pricing-model/","timestamp":"2014-04-18T19:28:31Z","content_type":null,"content_length":"43661","record_id":"<urn:uuid:34d02e3b-5c6f-4063-8a15-d5878010992b>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00592-ip-10-147-4-33.ec2.internal.warc.gz"} |
Class Summary
Class Description
Add This unit performs a signed addition on its two inputs.
AsymptoticRamp Output approaches Input exponentially.
ChannelIn Provides access to one channel of the audio input.
ChannelOut Provides access to one channel of the audio output.
Circuit Contains a list of units that are executed together.
Compare Output 1.0 if inputA > inputB.
ContinuousRamp A ramp whose function over time is continuous in value and in slope.
CrossFade CrossFade between parts of the input.
Delay Simple non-interpolating delay.
Divide This unit divides its two inputs.
DualInTwoOut This unit splits a dual (stereo) input to two discrete outputs.
EdgeDetector Output 1.0 if the input crosses from zero while rising.
EnvelopeAttackDecay Two stage Attack/Decay envelope that is triggered by an input level greater than THRESHOLD.
EnvelopeDAHDSR Six stage envelope similar to the envelope in DLS2.
ExponentialRamp Output approaches Input exponentially and will reach it eventually.
FFT Periodically transform the complex input signal using an FFT to a complex spectral stream.
FFTBase Periodically transform the complex input signal using an FFT to a complex spectral stream.
FilterBandPass Filter that allows frequencies around the center frequency to pass and blocks others.
FilterBandStop Filter that blocks frequencies around the center frequency.
FilterBiquad Base class for a set of IIR filters.
FilterBiquadCommon Extend this class to create a filter that implements Biquad filter with a Q port.
FilterBiquadShelf This filter is based on the BiQuad filter and is used as a base class for FilterLowShelf and FilterHighShelf.
FilterHighPass Filter that allows frequencies above the center frequency to pass.
FilterHighShelf HighShelf Filter.
FilterLowPass HighPass Filter.
FilterLowShelf LowShelf Filter.
FilterOnePole First Order, One Pole filter using the following formula:
FilterOnePoleOneZero First Order, One Pole, One Zero filter using the following formula:
FilterOneZero First Order, One Zero filter using the following formula:
FilterPeakingEQ PeakingEQ Filter.
FilterStateVariable A versatile filter described in Hal Chamberlain's "Musical Applications of MicroProcessors".
FilterTwoPoles Second Order, Two Pole filter using the following formula:
FilterTwoPolesTwoZeros Second Order, Two Pole, Two Zero filter using the following formula:
FixedRateMonoReader Simple sample player.
FixedRateMonoWriter Simple sample writer.
FixedRateStereoReader Simple stereo sample player.
FixedRateStereoWriter Simple stereo sample writer.
FourWayFade FourWayFade unit.
FunctionEvaluator Convert an input value to an output value.
FunctionOscillator Oscillator that uses a Function object to define the waveform.
Grain A single Grain that is normally created and controlled by a GrainFarm.
GrainFarm A unit generator that generates a cloud of sound using multiple Grains.
GrainSourceSine A simple sine wave generator for a Grain.
IFFT Periodically transform the complex input spectrum using an IFFT to a complex signal stream.
ImpulseOscillator Narrow impulse oscillator.
ImpulseOscillatorBL Impulse oscillator created by differentiating a sawtoothBL.
Integrate IntegrateUnit unit.
InterpolatingDelay InterpolatingDelay
Latch Latch or hold an input value.
LatchZeroCrossing Latches when input crosses zero.
LinearRamp Output approaches Input linearly.
LineIn External audio input is sent to the output of this unit.
LineOut Input audio is sent to the external audio output device.
Maximum Output largest of inputA or inputB.
Minimum Output smallest of inputA or inputB.
MonoStreamWriter Write one sample per audio frame to an AudioOutputStream with no interpolation.
Multiply This unit multiplies its two inputs.
MultiplyAdd output = (inputA * inputB) + inputC
Pan Pan unit.
PanControl PanControl unit.
ParabolicEnvelope ParabolicEnvelope unit.
PassThrough Pass the input through to the output unchanged.
PeakFollower Tracks the peaks of an input signal.
PinkNoise Random output with 3dB per octave rolloff providing a soft natural noise sound.
PitchDetector Estimate the fundamental frequency of a monophonic signal.
PowerOfTwo output = (2.0^input) This is useful for converting a pitch modulation value into a frequency scaler.
PulseOscillator Simple pulse wave oscillator.
PulseOscillatorBL Pulse oscillator that uses two band limited sawtooth oscillators.
RaisedCosineEnvelope An envelope that can be used in a GrainFarm to shape the amplitude of a Grain.
RectangularWindow Window that is just 1.0.
RedNoise RedNoise unit.
SampleGrainFarm A GrainFarm that uses a FloatSample as source material.
SawtoothOscillator Simple sawtooth oscillator.
SawtoothOscillatorBL Sawtooth oscillator that uses multiple wave tables for band limiting.
SawtoothOscillatorDPW Sawtooth DPW oscillator (a sawtooth with reduced aliasing).
SchmidtTrigger SchmidtTrigger unit.
Select SelectUnit unit.
SequentialDataReader Base class for reading a sample or envelope.
SequentialDataWriter Base class for writing to a sample.
SineOscillator Sine oscillator using fast Taylor expansion.
SineOscillatorPhaseModulated Sine oscillator with a phase modulation input.
SpectralFFT Periodically transform the input signal using an FFT.
SpectralIFFT Periodically transform the input signal using an Inverse FFT.
SquareOscillator Simple square wave oscillator.
SquareOscillatorBL Band-limited square wave oscillator.
StereoStreamWriter Write two samples per audio frame to an AudioOutputStream as interleaved samples.
StochasticGrainScheduler Use a random function to schedule grains.
Subtract This unit performs a signed subtraction on its two inputs.
TriangleOscillator Simple triangle wave oscillator.
TunableFilter A UnitFilter with a frequency port.
TwoInDualOut This unit combines two discrete inputs into a dual (stereo) output.
UnitBinaryOperator Base class for binary arithmetic operators like Add and Compare.
UnitFilter Base class for all filters.
UnitGate Base class for other envelopes.
UnitGenerator Base class for all unit generators.
UnitOscillator Base class for all oscillators.
UnitStreamWriter Base class for writing to an AudioOutputStream.
VariableRateMonoReader This reader can play any SequentialData and will interpolate between adjacent values.
VariableRateStereoReader This reader can play any SequentialData and will interpolate between adjacent values.
WhiteNoise WhiteNoise unit. | {"url":"http://www.softsynth.com/jsyn/docs/javadocs/com/jsyn/unitgen/package-summary.html","timestamp":"2014-04-18T21:42:03Z","content_type":null,"content_length":"33785","record_id":"<urn:uuid:15a1aba5-c3ac-4921-bc86-1b31e278a9d6>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00035-ip-10-147-4-33.ec2.internal.warc.gz"} |
SHORT TALK BULLETIN - Vol.XIII March, 1935 No.3
THAT ANCIENT SQUARE
by: Unknown
What one symbol is most typical of Freemasonry as a whole? Mason and non-Mason alike, nine times out of ten, will answer, “The Square!” Many learned writers on Freemasonry have denominated the square
as the most important and vital, most typical and common symbol of the ancient Craft. Mackey terms it “one of the most important and significant symbols.” McBride said:
“-In Masonry or building, the great dominant law is the law of the square.” Newton’s words glow: “Very early the square became an emblem of truth, justice and righteousness, and so it remains to this
day, though uncountable ages have passed. Simple, familiar, eloquent; it brings from afar a sense of wonder of the dawn, and it still teaches a lesson we find it hard to learn.” Haywood speaks of:
“—Its history, so varied and so ancient, its use, so universal.”
“An important emblem - passed into universal acceptance.” In his encyclopedia, Kenning copied Mackey’s phrase. Klein reverently denominates it “The Great Symbol.” I Kings, describing the Temple,
states that “all the doors and the posts were square.” It is impossible definitely to say that the square is the oldest symbol in Freemasonry; who may determine when the circle, triangle or square
first impressed men’s minds? But the square is older than history. Newton speaks of the oldest building known to man: “- A prehistoric tomb found in the sands at Hieraconpolis, is already right
Masonically the word “square” has the same three meanings given the syllable by the world: (1) The conception of right angleness - our ritual tells us that the square is an angle of ninety degrees,
or the fourth of a circle; (2) The builder’s tool, one of our working tools, the Master’s own immovable jewel; (3) That quality of character which has made “a square man” synonymous not only with a
member of our Fraternity, but with uprightness, honesty and dependability.
The earliest of the three meanings must have been the mathematical conception. As the French say, “it makes us furiously to think” to reflect upon the wisdom and reasoning powers of men who lived
five thousand years ago, that they knew the principles of geometry by which a square can be constructed.
Plato, greatest of the Greek philosophers, wrote over the porch of the house in which he taught: “Let no one who is ignorant of geometry entry my doors.” Zenocrates , a follower of Plato, turned away
an applicant for the teaching of the Academy, who was ignorant of geometry, with the words: “Depart, for thou has not the grip of philosophy.” Geometry is so intimately interwoven with architecture
and building that “geometry, or Masonry, originally synonymous terms” is a part of most rituals. The science of measurements is concerned with angles, the construction of figures, the solution of
problems concerning both, and all the rest upon the construction of a right angle, the solutions which sprang from the Pythagorean Problem, our “Forty-Seventh Problem of Euclid,” so prominent in the
Master’s Degree.
The ancient Greek name of the square was “gnomon,” from whence comes our word “knowledge.” The Greek letter “gamma” formed like a square standing on one leg, the other pointing to the right - in all
probability derived from the square, and “gnomon,” in turn, derived from the square which the philosophers knew was at the root of their mathematics.
Democritus, old philosopher, according to Clement of Alexandria, once exulted: “In the construction of plane figures with proof, no one has yet surpassed me, not even the Harpedonaptae of Egypt.” In
the truth of his boast we have no interest, but much in the Harpedonaptae of Egypt. The names means, literally, “rope stretchers” or “Rope fasteners.” In the Berlin museum is a deed, written on
leather, dating back to 2,000 B.C. which speaks of the work of rope stretchers; how much older rope stretching may be, as a means of constructing a square, is unknown, although the earliest known
mathematical hand-book (that of Ahmes, who lived in the sixteenth or seventeenth Hyskos dynasty in Egypt, and is apparently a copy of a much older work which scholars trace back to 3400 B.C.), does
not mention rope stretching as a means of square construction. Most students in school days learned a dozen ways of erecting one line perpendicular to another. It seems strange that any other people
were ever ignorant of such simple mathematics. Yet all knowledge had a beginning. Masons learn of Pythagorean’s astonishment and delight at his discovery of the principle of the Forty-seventh
Problem. Doubtless the first man who erected a square by stretching a rope was equally happy over his discovery. Researchers into the manner of construction of pyramids, temples and monuments in
Egypt reveal a very strong feeling on the part of the builders for the proper orientation of their structures. Successfully to place the building so that certain points, corners or openings might
face the sun or a star at a particular time, required very exact measurements. Among these, the laying down of the cross axis at a right angle to the main axis of the structure was highly important.
It was this which the Harpedonaptae accomplished with a long rope. The cord was first marked off in twelve equal portions, possible by knots, more probably, by markers thrust into the body of the
rope. The marked rope was then laid upon the line on which a perpendicular (right angle) was to be erected. The rope was pegged down at the third marker from the from one end, and another, four
markers further on. This left two free ends, one three total parts long, one five total parts long. With these ends the Harpedonatae scribed two semi-circles. When the point where these two met, was
connected to the first peg (three parts from the end of the rope, a perfect right angle, or square, resulted.
Authorities have differed and much discussion has been had, on the “true form” of the Masonic square; whether a simple square should be made with legs of equal length, and marked with divisions into
feet and inches, or with one keg longer than the other and marked as are carpenter’s squares today. Mackey says:
“It is proper that its true form should be preserved. The French Masons have almost universally given it with one leg longer than the other, thus making it a carpenter’s square. The American Masons,
following the delineations of Jeremy L. Cross, have, while generally preserving the equality of length in the legs, unnecessarily marked its surface with inches, thus making it an instrument for
measuring length and breadth, which it is not. It is simply the “trying square” of a stonemason, and has a plain surface, the sides embracing an angle of ninety degrees, and it is intended only to
test the accuracy of the sides of a stone, and to see that its edges subtend the same angle.”
Commenting on this, the Editor of “the Builder” wrote (May, 1928):
“This is one of the occasions when this eminent student ventured into a field beyond his own knowledge, and attempted to decide a matter of fact from insufficient data. For actually, there is not,
and never has been, any essential difference between the squares used by carpenters and stone workers. At least not such difference as Mackey assumes. He seems to imply that French Masons were guilty
of an innovation in making the square with unequal limbs. This is rather funny, because the French (and the Masons of Europe generally) have merely maintained the original form, while English
speaking Masonry, or rather the designers of Masonic jewels and furnishings in English speaking countries, have introduced a new form for the sake, apparently, of its greater symmetry. From medieval
times up till the end of the eighteenth century, all representations of Mason’s squares show one limb longer than the other. In looking over the series of Masonic designs of different dates it is
possible to observe the gradual lengthening of the shorter limb and the shortening of the longer one, till it is sometimes difficult to be certain at first glance if there is any difference between
them. “There is absolute no difference in the use of the square in different crafts. In all the square is used to test work, but also to set it out. And a square with a graduated scale on it is at
times just as great a convenience for the stonemason as for the carpenter. When workmen made their own squares there would be no uniformity in size or proportions, and very few would be graduated,
though apparently this was sometimes done. It is rather curious that the cut which illustrates this article in Mackey’s Encyclopedia actually show a square with one limb longer than the other.” It is
to be noted that old operative squares were either made wholly of wood, or of wood and metal, as indeed, small try squares are made today. Having one leg shorter than the other would materially
reduce the chance of accident destroying the right angle which was the tools essential quality . . So that authorities who believe our equal legged squares not necessarily “true Masonic squares” have
some practical reasons for their convictions.
It is of interest to recall McBride’s explanation of the “center” as used in English Lodges, and the “point within a circle,” familiar to us. He traces the medieval “secret of the square” to the use
of the compasses to make the circle from which the square is laid out.
Lines connecting a point, placed anywhere on the circumference of a
circle, to the intersection with the circumference cut by a straight
line passing through the center of the circle, forms a perfect
square. McBride believed that our “point within a circle” was direct
reference to this early operative method of correcting the angles in
the wooden squares of operative cathedral builders, and that our present “two perpendicular lines” are a corruption of the two lines which connect points on the circle.
The symbolism of the square, as we know it, is also very old; just how ancient, as impossible to say as the age of the tool or the first conception of mathematical “square-ness.” In 1880 the Master
of Ionic Lodge No. 1781, at Amot, China, speaking on Freemasonry in China said:
“From time immemorial we find the square and compasses used by Chinese writers to symbolize precisely the same phrases of moral conduct as in our system of Freemasonry. The earliest passage known to
me which bears upon the subject is to be found in the Book of History embracing the period reaching from the twenty-fourth to the seventh century before Christ. There is an account of a military
expedition where we read:
“Ye Officers of government, apply the Compasses!” “In another part of the same venerable record a Magistrate is spoken of as: ‘A man of the level, or the level man.’ “The public discourses of
Confucius provide us with several Masonic allusions of a more or less definite character. For instance, when recounting his own degrees of moral progress in life, the Master tells us that only at
seventy-five years of age could he venture to follow the inclinations of his heart without fear of ‘transgressing the limits of the square.’ This would be 481 B.C., but it is in the words of the
great follower, Mencius, who flourished nearly two hundred years later, that we meet with a fuller and more impressive Masonic phraseology. In one chapter we are taught that just as the most skilled
articifers are unable, without the aid of the square and compasses, to produce perfect rectangles or perfect circles, so must all men apply these tools figuratively to their lives, and the level and
the markingline besides, if they would walk in the straight and even paths of wisdom, and keep themselves within the bounds of honor and virtue. In Book IV we read:
“The compasses and Square are the embodiment of the rectangular and the round, just as the prophets of old were the embodiment of the due relationship between man and man.”
In Book IV we find these words:
“The Master Mason, in teaching his apprentices, makes use of the compasses and the square. Ye who are engaged in the pursuit of wisdom must also make use of the compasses and the square.” In the
“Great Learning,” admitted on all sides to date from between 300 to 400 years before Christ, in Chapter 10, we read that a man should abstain from doing unto others what he would not they should do
unto him: “this,” adds the writer, “is called the principle of acting on the square.”
Independently of the Chinese, all peoples in all ages have thought of this fundamental angle, on which depends the solidity and lasting quality of buildings, as expressive of the virtues of honesty,
uprightness and morality. Confucius, Plato, the Man of Galilee, stating the Golden Rule in positive form, all make the square an emblem of virtue.
In this very antiquity of the Craft’s greatest symbol is a deep lesson; the nature of a square is as unchanging as truth itself. It was always so, it will always be so. So, also, are those principles
of mind and character symbolized by the square; the tenets of the builder’s guild expressed by a square. They have always been so, they will always be so. From their very nature they must ring as
true on the farthest star as here.
So will Freemasonry always read it, that its gentle message perish not from the earth! | {"url":"http://www.masonicworld.com/education/files/artjuly01/thatancientsquare.htm","timestamp":"2014-04-21T07:04:47Z","content_type":null,"content_length":"20752","record_id":"<urn:uuid:0ed70ba9-b386-4fa7-bf0f-b5923cf0782a>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00653-ip-10-147-4-33.ec2.internal.warc.gz"} |
Departmental Awards
The Giovanni Borrelli Mathematics Fellowship
The Mathematics Department is aware of the fact that many HMC mathematics majors have the ability to complete high-quality research, either independently or in tandem with a faculty advisor. The
Giovanni Borrelli Fellowship was created to confer some degree of distinction upon such qualified students as well as to allow them an opportunity to become more intimately involved with scholastic
pursuits. Giovanni Borrelli Fellows are expected to pursue research in the year following the award, in particular over the ensuing summer. They may also be called upon to represent the Mathematics
Department at various events, such as entertaining invited speakers or serving as student representatives on search committees.
Winners of the Borrelli Fellowship.
The Giovanni Borrelli Mathematics Prize
The Giovanni Borrelli Mathematics Prize is an annual award presented to rising senior mathematics students who have distinguished themselves scholastically and show promise of contributing in an
important way to the mathematical sciences. This award is presented in the fall and is accompanied by a cash prize. It is normally not awarded to a winner of the Giovanni Borrelli Mathematics
Winners of the Borrelli Prize.
The Stavros Busenberg Prize in Applied Mathematics
The Stavros Busenberg Prize in Applied Mathematics is an annual award presented to rising senior students who show particular promise in the study of applied mathematics. This award is presented in
the fall and is accompanied by a cash prize. It normally is not awarded to a recipient of the Giovanni Borrelli Mathematics Prize or the Giovanni Borrelli Mathematics Fellowship.
Winners of the Busenberg Prize.
Henry A. Krieger Prize in Decision Sciences
The Henry A. Krieger Prize in Decision Sciences is an annual award presented to rising senior students who show particular promise in probability, statistics, or operations research This award is
presented in the fall and is accompanied by a cash prize. It is normally not awarded to a recipient of the Stavros Busenberg Price, the Giovanni Borrelli Mathematics Prize, or the Giovanni Borrelli
Mathematics Fellowship.
The Courtney S. Coleman Prize (Sophomore Mathematics Award)
The Courtney S. Coleman Prize is given to two rising juniors who excel in mathematics. The award is presented at the HMC convocation in the fall.
Winners of the Courtney S. Coleman Prize.
The Robert James Prize (Freshman Mathematics Award)
The Robert James Prize is given to two rising sophomores who excel in mathematics. The award is presented at the HMC convocation in the fall.
Winners of the Robert James Prize.
The Alvin White Prize
The Alvin White Prize is an annual award presented to an HMC student or students who have contributed greatly to the humanistic side of the Harvey Mudd College Mathematics Community and beyond. The
award recognizes contributions to making mathematics more accessible and enjoyable, enriching the mathematical experiences at the College, performing mathematical education outreach, and/or the use
of mathematics in the service and betterment of humanity. These values embody the spirit of Alvin White's vision of a kinder, gentler and more humanistic approach to the practice and teaching of
Winners of the Alvin White Prize.
The Chavin Prize
The Chavin Prize is an annual award to HMC students who author papers in the area of the mathematical sciences. This award is presented at the HMC senior awards ceremony in the spring, and is
accompanied by a cash prize.
The Greever Clinic Award
The Greever Award is awarded annually to students (individually or as a team) who make an outstanding contribution as part of the HMC Mathematics Clinic program. The award is named in honor of Prof.
John Greever who played a leading role in the founding of the Mathematics Clinic program in 1973.
The RIF Scholarship
For a description of the RIF scholarship, see the RIF Scholarship page.
Recipients of the RIF scholarship.
The RIF Prizes
The RIF Prizes are cash prizes awarded to the top three HMC students on the William Lowell Putnam Mathematical Competition.
The Jayaweera Prize
The Jayaweera Prize is awarded for outstanding performance in the annual Interdisciplinary Competition in Modeling. | {"url":"http://www.math.hmc.edu/program/awards/","timestamp":"2014-04-20T23:29:52Z","content_type":null,"content_length":"11948","record_id":"<urn:uuid:da88d457-ffc0-4b8c-bc1d-49ae91d1b78c>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00623-ip-10-147-4-33.ec2.internal.warc.gz"} |
Need matlab help!!!
January 29th 2010, 09:47 AM
Need matlab help!!!
Hi i'm taking an introductory course to matlab at Fau and i'm having a little trouble with my first project. Seeing as how this looks completely foreign to me, I have no clue what to do actually.
I was wondering if someone could help me out. This is the project template:
Develop a computer program in MATLAB that will evaluate the following function for -0.9 ≤ x ≤ 0.9 in steps of 0.1 by:
A) an arithmetic statement .
B) by series allowing for as many as 50 terms. However, end adding terms when the last term only affects the 6th place in the answer.
The function and its series expansion is:
f(x)= (1+x^2)^.5= 1-1/2x^2 + 1/2 x 3/4x^4 - 1/2 x 3/4 x 5/6x^6 +-....
Print out a table (to a file) in the following format; use 6 decimal places for f(x).
X f(x)(by arith stm) f(x)(by series) ]
- ...
- ...
and there should be one last row that says # of terms used in the series
This program should be written off of these examples...
Example 2.3
% while1.m
% Calculation of e^x by a Taylor series using a while loop. The 'input
% statement' is used to establish the exponent, x. A 'while loop' is
% used in determining the series solution. In this example term(n) is
% obtained by multiplying term(n-1) by x and dividing by the index n.
clear; clc;
x=input('enter a value for the exponent x \n');
while abs(term) > ex*1.0e-6
if n > 50
disp(x); disp(ex);
EXAMPLE 2.2
% exB.m
% The program calculates of e^x by both an arithmetic statement (ex2)
% and by a Taylor series (ex1), where -0.5< x <0.5. A 'for loop' is
% used to determine the x values. In the series part of the program,
% the 'sum' function is used to sum all the terms calculated in the
% inner 'for loop'. Fifty terms are used in the series.
clear; clc;
xmin=-0.5; dx=0.1;
% Table headings
fprintf(' x ex1 ex2 \n');
for i=1:11
for n=1:50
fprintf('%5.2f %10.5f %10.5f \n',x,ex1,ex2);
If someone could at least tell me how to go about starting to write this program. I would be very appreciative.
P.s. I tried to get the appropriate spacings to translate over to this text box but it just wasn't having it.
January 29th 2010, 10:51 PM
Hi i'm taking an introductory course to matlab at Fau and i'm having a little trouble with my first project. Seeing as how this looks completely foreign to me, I have no clue what to do actually.
I was wondering if someone could help me out. This is the project template:
Develop a computer program in MATLAB that will evaluate the following function for -0.9 ≤ x ≤ 0.9 in steps of 0.1 by:
A) an arithmetic statement .
B) by series allowing for as many as 50 terms. However, end adding terms when the last term only affects the 6th place in the answer.
The function and its series expansion is:
f(x)= (1+x^2)^.5= 1-1/2x^2 + 1/2 x 3/4x^4 - 1/2 x 3/4 x 5/6x^6 +-....
Print out a table (to a file) in the following format; use 6 decimal places for f(x).
X f(x)(by arith stm) f(x)(by series) ]
- ...
- ...
and there should be one last row that says # of terms used in the series
This program should be written off of these examples...
Example 2.3
% while1.m
% Calculation of e^x by a Taylor series using a while loop. The 'input
% statement' is used to establish the exponent, x. A 'while loop' is
% used in determining the series solution. In this example term(n) is
% obtained by multiplying term(n-1) by x and dividing by the index n.
clear; clc;
x=input('enter a value for the exponent x \n');
while abs(term) > ex*1.0e-6
if n > 50
disp(x); disp(ex);
EXAMPLE 2.2
% exB.m
% The program calculates of e^x by both an arithmetic statement (ex2)
% and by a Taylor series (ex1), where -0.5< x <0.5. A 'for loop' is
% used to determine the x values. In the series part of the program,
% the 'sum' function is used to sum all the terms calculated in the
% inner 'for loop'. Fifty terms are used in the series.
clear; clc;
xmin=-0.5; dx=0.1;
% Table headings
fprintf(' x ex1 ex2 \n');
for i=1:11
for n=1:50
fprintf('%5.2f %10.5f %10.5f \n',x,ex1,ex2);
If someone could at least tell me how to go about starting to write this program. I would be very appreciative.
P.s. I tried to get the appropriate spacings to translate over to this text box but it just wasn't having it.
Look at the examples. Load them and single step through in the debugger to get an idea of how they work.
Then change the calculation of the next term in the series from that in the examples to that for the function you have been asked for.
February 1st 2010, 06:52 PM
This is pretty much what i was able to write on my own today by comparing what i needed to result in and what the examples show.
clc; clear;
xmin=-.9; dx=.1;
fprintf(' x f(x)(my arith. statement) f(x)(by series) # of terms per series \n');
fprintf('-------------------------------------------------------------- /n');
for i=1:19
for n=1:50
if (abs(term) <= 1.00e-6;
fprintf('%5.2f %10.6f %10.6f \n',x,ex2,ex1);
I don't know if i did this right... i'm not even sure if it's putting out the results my project is asking me to project... I also know that i need to "print" my results to a file which includes
something like --- fid=fopen('output.dat','w'); and so on... but i'm not sure how to incorporate that. I also don't know how to set it up so i list the number of terms/ series in the last column.
That part is completly beyond me too. My book is absolutely no help... I wish it was. but i've read the chapter three tiems and i'm still lost. HELP!! | {"url":"http://mathhelpforum.com/math-software/126152-need-matlab-help-print.html","timestamp":"2014-04-19T10:52:06Z","content_type":null,"content_length":"12804","record_id":"<urn:uuid:b9e7ab8e-2c6f-4550-90ba-29ec78151a0c>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00155-ip-10-147-4-33.ec2.internal.warc.gz"} |
Inkscape tutorial: Interpolar
Inkscape tutorial: Interpolar
Ryan Lerch, ryanlerch arroba gmail punto com
Este documento explica como usar a extensión de interpolación de Inkscape
Interpolate does a linear interpolation between two or more selected paths. It basically means that it “fills in the gaps” between the paths and transforms them according to the number of steps
To use the Interpolate extension, select the paths that you wish to transform, and choose from the menu.
Before invoking the extension, the objects that you are going to transform need to be paths. This is done by selecting the object and using or Shift+Ctrl+C. If your objects are not paths, the
extension will do nothing.
Interpolation between two identical paths
The simplest use of the Interpolate extension is to interpolate between two paths that are identical. When the extension is called, the result is that the space between the two paths is filled with
duplicates of the original paths. The number of steps defines how many of these duplicates are placed.
Por exemplo vexamos estes dous camiños:
Now, select the two paths, and run the Interpolate extension with the settings shown in the following image.
As can be seen from the above result, the space between the two circle-shaped paths has been filled with 6 (the number of interpolation steps) other circle-shaped paths. Also note that the extension
groups these shapes together.
Interpolación entre dous camiños diferentes
When interpolation is done on two different paths, the program interpolates the shape of the path from one into the other. The result is that you get a morphing sequence between the paths, with the
regularity still defined by the Interpolation Steps value.
Por exemplo vexamos estes dous camiños:
Now, select the two paths, and run the Interpolate extension. The result should be like this:
As can be seen from the above result, the space between the circle-shaped path and the triangle-shaped path has been filled with 6 paths that progress in shape from one path to the other.
When using the Interpolate extension on two different paths, the position of the starting node of each path is important. To find the starting node of a path, select the path, then choose the Node
Tool so that the nodes appear and press TAB. The first node that is selected is the starting node of that path.
See the image below, which is identical to the previous example, apart from the node points being displayed. The node that is green on each path is the starting node.
The previous example (shown again below) was done with these nodes being the starting node.
Now, notice the changes in the interpolation result when the triangle path is mirrored so the starting node is in a different position:
Método de interpolación
One of the parameters of the Interpolate extension is the Interpolation Method. There are 2 interpolation methods implemented, and they differ in the way that they calculate the curves of the new
shapes. The choices are either Interpolation Method 1 or 2.
Nos exemplos de arriba usouse o Método de interpolación 2 e o resultado foi:
Compare agora isto co Método de interpolación 1:
The differences in how these methods calculate the numbers is beyond the scope of this document, so simply try both, and use which ever one gives the result closest to what you intend.
The exponent parameter controls the spacing between steps of the interpolation. An exponent of 0 makes the spacing between the copies all even.
Here is the result of another basic example with an exponent of 0.
O mesmo exemplo cando o expoñente é 1:
cando o expoñente é 2:
e cando o expoñente é -1:
When dealing with exponents in the Interpolate extension, the order that you select the objects is important. In the examples above, the star-shaped path on the left was selected first, and the
hexagon-shaped path on the right was selected second.
View the result when the path on the right was selected first. The exponent in this example was set to 1:
Duplicate Endpaths
This parameter defines whether the group of paths that is generated by the extension includes a copy of the original paths that interpolate was applied on.
Estilo de interpolación
This parameter is one of the neat functions of the interpolate extension. It tells the extension to attempt to change the style of the paths at each step. So if the start and end paths are different
colors, the paths that are generated will incrementally change as well.
Aquí ten un exemplo de como se usa a función Estilo de interpolación no recheo dun camiño:
O estilo de interpolación tamén lle afecta ao trazo dun camiño:
Of course, the path of the start point and the end point does not have to be the same either:
Using Interpolate to fake irregular-shaped gradients
It is not possible in Inkscape (yet) to create a gradient other than linear (straight line) or radial (round). However, it can be faked using the Interpolate extension and Interpolate Style. A simple
example follows — draw two lines of different strokes:
And interpolate between the two lines to create your gradient:
As demonstrated above, the Inkscape Interpolate extension is a powerful tool. This tutorial covers the basics of this extension, but experimentation is the key to exploring interpolation further. | {"url":"http://www.inkscape.org/doc/interpolate/tutorial-interpolate.gl.html","timestamp":"2014-04-18T18:42:46Z","content_type":null,"content_length":"15175","record_id":"<urn:uuid:2c7f83f4-bf86-45c2-92bf-46fd87ad2c89>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00382-ip-10-147-4-33.ec2.internal.warc.gz"} |
After all the fun I had making fractal snowflakes, I wanted to try out some different ideas with the “Transform Effect” in Illustrator. Here I typed the word “Dragon,” and applied an effect that made
three copies, rotated in multiples of 42º (I went for an arbitrary angle other than 60), and scaled up and shifted a little each time.
Applying the same effect again makes a kind of spiral, as you might imagine.
But because the rotations are done with respect to the bottom-left corner (rather than the center of the spiral), a strange pattern begins to emerge:
Here I tried it again, with a transformation that makes one copy rotated 80º about the lower left corner, and shifted right.
We might expect to see four “dragon”s when the transformation is applied again, but one of them falls perfectly on the first copy so we see only three.
After the third iteration, the dragons have curled around enough to be farther left than the original, which moves the point of rotation. Soon the fractal magic begins to occur.
Watching the movie Frozen with my kids the other day, I was happy to hear the term “frozen fractals” in the acclaimed song, “Let It Go.” It made me look closer at the snowflakes and ice shapes
depicted in the movie – but I didn’t notice much in the way of fractal structure there.
The archetypical snowflake images tend to exhibit “branches on branches,” indicating fractal self-similarity, though the six-fold symmetry appears to be the defining characteristic of iconic
snowflake shapes.
I wondered about making my own fractal snowflakes using a simple iterative process…but I wanted something different than the Koch snowflake.
Starting with a random shape, I repeated the following process (in Adobe Illustrator, which made it easy):
Create 5 more copies, rotated in multiples of 60º about the lower left corner of the bounding box.
Here are the first four iterations:
After that my computer was slowing down, but you get the idea. I tried the same idea again, but this time the copies were rotated about one vertex so they overlapped and ended up making a cool tile
Here’s a caricature I made of “the man who loved only numbers” for his 101st birthday.
Happy birthday, Albert!
Here’s a fun site for your lateral thinking students (who also know some geometry). In their words,
A team of logicians adapted or created these puzzles - some require research, some require mathematics, some require pure savvy.
A lot of the problems don’t have anything to do with pi, and some of the presentation is a little janky, but it’s still fun.
I recently came across some cool ideas for blending up art and math for younger kids at whatdowedoallday.com – just in time for Pi Day next Friday!
I came across this picture I took while visiting the bell tower of the Old Post Office in Washington, D.C. a couple of years ago. It describes how the bells are occasionally rung in a “full peal,”
consisting of all possible permutations of their tones. With 6 bells, you get 6! = 6*5*4*3*2*1 = 720 permutations, which could be rung in under an hour, but with 7 bells, you get 7! = 7*720 = 5040
patterns to ring!
I thought maybe this would involve listing the set of permutations in increasing numerical order:
But In the example they showed, it seems that the pattern is to first swap neighbors in 3 groups:
And then to keep the first and last positions in place but swap neighbors in the 2 groups inside:
I can’t tell if that’s all there is to the method or not, and I wonder if there’s a proof that the method produces all permutations. A quick test with 4 bells shows that you don’t get all 24
permutations before it repeats. | {"url":"http://www.10minutemath.com/?p=84","timestamp":"2014-04-20T01:18:20Z","content_type":null,"content_length":"138486","record_id":"<urn:uuid:11e6dde9-63aa-4417-ad14-fae8db0991b9>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00455-ip-10-147-4-33.ec2.internal.warc.gz"} |
Area of Squares
Re: Area of Squares
oooh 10 by 10 so 100
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
Got it. But let's improve the accuracy.
See enlarged diagram. It's not quite the whole of a square so take a bit off 100.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Area of Squares
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
OK. I'll accept that.
Now to put everything together. You will need to substitute a value for PI in your earlier answers.
Then you can do yellow + red + green less 95 = ???
Put down all the figures so I can check each bit of the calculation.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Area of Squares
OK just before I do that So 225PI - 95 = 611.858347058
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
Yes, that calculation is correct. But where did 225 come from.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Area of Squares
the 1/4 the green part area
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
But the radius is not 15.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Area of Squares
the radius is 30 right ?
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
Now you have all the figures individually so it's time to put them together. The goat is getting hungry.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Area of Squares
3/4 + 50²+30²+30² * Pi = 5520.575041
1875pi + 225 Pi + 2401 Pi = 4501 Pi =14140.3085338
Last edited by zee-f (2012-11-16 22:00:04)
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
hi zee-f
Oh dear. A bit of a muddle. You had the correct yellow area back in post 81.
Also remember in post 79 you had a good estimate of the answer by counting squares. I said then, I thought it was a bit high. So you should not be satisfied untill you land up with an answer that is
somewhere between these two.
Let me set out for you what you should be calculating.
step 1. area of yellow quadrants = 3/4 x 50^2 x PI =
step 2. area of red quadrant = 1/4 x 30^2 x PI =
step 3. area of green quadrant = same answer =
step 4 estimated area (overlap) = 95
step 5. add up 1, 2 and 3 and take away 4 total =
Please set it out like this then, if there is still a gremlin lurking in there, I can see where it is.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Area of Squares
3/4 x 50^2 x π + 1/4 x 30^2 x π + 1/4 x 30^2 x π =
1875π + 225π + 225π= 2325π
2325PI - 95 = 7209.20292
Last edited by zee-f (2012-11-17 00:22:57)
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
That's what I've got too.
Well done.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Area of Squares
as you have a whole number estimate in the calculation it would be sensible to round off the answer to, say, the nearest hundred.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Area of Squares
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
9. The areas you found in 7 and 8 overlap each other. How much do they overlap? What *approximate* shape do they make? What is that area?
about 250 Looks like 3 triangles uniting together the area is
3/4 x 50^2 x π + 1/4 x 30^2 x π + 1/4 x 30^2 x π =
1875π + 225π + 225π= 2325π
2325PI - 250= 7054.20292
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
10. What is the total grazing area the goat can reach?
3/4 +1/4= 4/4
50^2 + 30^2 = 80^2
4/4 * 80 ^2 * PI = 6400 PI
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
I thought we had already reached Q10. That's what you've been working towards.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Area of Squares
6. How much of the 50 foot circle can the goat reach without getting interrupted by the barn? What is that area?
6. 3/4 * 50^2 * PI= 1875 PI
7. When the rope goes around the barn, what is the new radius? How much of a circle can it make without hitting the barn or overlapping area you've already found? What is that area?
7- 30 is the new radius. It can make 1/4. The area is 1/4 * 30² * PI = 225 PI
8. When the rope goes around the barn the other way, what is the new radius? How much of a circle can it make without hitting the barn or overlapping area you've already found? What is that area?
8. the radius is 30. It can make 1/4 the area is 1/4 * 30² * PI = 225 PI
9. The areas you found in 7 and 8 overlap each other. How much do they overlap? What *approximate* shape do they make? What is that area?
9. 95. They look like 3 triangles uniting. the area is
(3/4 * 50^2 * PI= 1875 PI + 1/4 * 30² * PI = 225 PI + 1/4 * 30² * PI = 225 PI) - 95
2325PI - 95 = 7209.2020292
10. What is the total grazing area the goat can reach?
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
oooh yeah they overlap about 250 So the answer I put for #9 is number 10 right?
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
Yes that looks good to me.
Your answer for Q9 is a little less than one full square, which you estimate as 95.
Then the long calculation is for Q10.
The question for Q8 doesn't actually make sense so to cover all possible angles on this I would answer it a bit like this:
8. When the rope goes around the barn the other way, what is the new radius?
How much of a circle can it make without hitting the barn or overlapping area you've already found? What is that area?
Initially the radius is 50 again; then dropping to 30 and finally to 10. The answers are exactly the same as when the goat went round the first way.
The 3/4 part (radius = 50) is the same as before and so the non-overlapping bit for this is zero.
The 1/4 part (radius = 30) is the same but it does overlap a little with the answer already obtained.
If what I've said makes sense to you, I would prefer that you put it into your own words rather than just copy me. That way you can hold your head up and say you did this rather than being accused of
just copying.
I think that is the end of the whole thing. Hurrah!!
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Area of Squares
Am ok with my answer for #8 but am confused between #9 # 10 now So
(3/4 * 50^2 * PI= 1875 PI + 1/4 * 30² * PI = 225 PI + 1/4 * 30² * PI = 225 PI) - 95
2325PI - 95 = 7209.2020292
#9- 9. The areas you found in 7 and 8 overlap each other. How much do they overlap? What *approximate* shape do they make? What is that area?
They overlap 95 but what's the area here?
One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3
Re: Area of Squares
Q 9. The approximate shape is a square. As it's not quite a whole square, you estimated it's area at about 95 square feet.
Q10. Is spot on. But round it off. (i) because part of the calculation is an estimate so 7 decimal places is way over the top and (ii) the calculation is to say what area of grass the goat can reach.
So how accurate do you need to be anyway? The actual amount the goat gets will depend on other factors like the quality of the grass (it may be better in one place than another) and how far the goat
is prepared to stretch to reach that deliciously succulent bit etc etc. 7 decimal places would amount to tiny fractions of a single blade of grass.
I'm going to log out for a while. I'll check in again later on.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Area of Squares
I completed the lesson got a ten thank you sooooo much
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How Many Trees?
September 12th 2010, 09:52 PM #1
Junior Member
Aug 2010
How Many Trees?
A forest contains 843,000 trees of a certain species. Statistical studies show that the number of leaves on these trees varies from a low of 943 to a high of 3722. At least how many trees must
have the same number of leaves?
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Geometric motivation for the Stanley-Reisner correspondence
up vote 8 down vote favorite
The Stanley-Reisner ring of an abstract simplicial complex $\Delta$ on the vertex set $\{1,...,n\}$ is the $k$-algebra $$ k[X_1,...,X_n]/I_\Delta $$ where $I_\Delta$ is the ideal generated by the $X_
{i_1}...X_{i_r}$ with ${i_1,...,i_r}\notin \Delta$.
Somebody told me that this construction helps to study varieties $k[X_1,...,X_n]/I$ for an arbitrary ideal $I$ as follows (if I am not missing something): Let $X_1<...< X_n$ be a monomial order and
consider the initial ideal $I':=in_{<}(I)$ of $I$. The passage from $k[X_1,...,X_n]/I$ to $k[X_1,...,X_n]/I'$ is called 'flat deformation' and this term makes sense if I draw pictures of the
varieties. Many properties of $I$ (like dimension) are directly related to properties of $I'$.
The aim is to find a to $I'$ related ideal $I_\Delta$ for an abstract simplicial complex $\Delta$.
I was told that the problem that $I'$ has a generator like $X_1^2$ could be resolved by introducing a new variable $X_1'$, replacing $X_1X_1$ by $X_1X_1'$ and mod out $X_1-X_1'$ of $k
[X_1,...,X_n,X_1']$. First, I don't understand why this should be closer to the form $I_\Delta$ because $I_\Delta$ is generated by monomials.
My main question is:
Can one see in a concrete affine example how the geometry of $\Delta$ relates to the initial variety $I$?
If you take for example the simplicial complex $\Delta$ (I apologize that I can not typeset the brackets) $\emptyset,X, Y, Z, XY, XZ, YZ$ which looks like a one sphere, the associated variety is the
union of the $XY$, the $XZ$ and the $YZ$ hyperplane in $\mathbb{R}[X,Y,Z]$. Why is this reasonable? On the other hand, I would like to know how the simplicial complex (after the transformation
indicated above) corresponding to the circle $I=(X^2+Y^2-1)$ in $\mathbb{R}[X,Y,Z]$ looks like. It would be nice if this is associated to the $\Delta$.
Please tell me, if this question is completely unreasonable or doesn't make any sense. I am an absolute beginner in algebraic geometry.
ag.algebraic-geometry ac.commutative-algebra co.combinatorics
Edited title to avoid confusion about the name Reisner. – Jim Humphreys Dec 9 '10 at 23:12
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2 Answers
active oldest votes
First when it comes to comparison with the simplicial complex it should be realised that the Stanley-Reisner ring corresponds to the cone over the complex. There is a non-homogeneous
version of it where one replaces the linear subspace through the origin by the affine space parallel to the linear space but passing through $1$ (the way the standard simplices are
defined). If the simplicial complex is a cone with a fixed apex, the Stanley-Reisner ring of the base of the cone is isomorphic to the non-homogeneous ring of the cone. (Incidentally, or
maybe not so incidentally, for a field of characteristic zero this non-homogeneous ring is the degree zero part of the Sullivan differential graded algebra associated to the complex.)
The Stanley-Reisner rings also pop back up for a general complex as being related to the stars of vertices, the stars being cones with apex the point. More precisely the localisation of
the Stanley-Reisner ring at the irrelevant maximal ideal is equal to the localisation of the non-homogeneous ring at the point. As the Stanley-Reisner ring is graded this localisation is
"essentially" the same as the graded Stanley-Reisner ring. This is why the properties of the Stanley-Reisner ring reflect the local properties of the complex, i.e., of the links of
vertices as well as the global properties as the link of the apex of the cone of a complex is the complex itself.
If the base field is the real numbers, then the space of real points of the non-homogeneous ring of a complex is homotopy equivalent to the complex itself. Indeed, the real points is the
union of the affine span of the simplices of the complex and one may retract each such affine space to the simplex in a way which is compatible with passing to the subsimplices.
up vote 4
down vote Finally, as for the flat deformation an arbitrary ideal gives rise to a monomial ideal, an ideal generated by monomials. The monomial ideals associated to the a simplicial complex are
accepted special monomial ideals; they are exactly those monomial ideals generated by square free monomials. Depending on the ordering of the variables the monomial ideal associated to $X^2+Y^2-1$
is $(X^2)$ or $(Y^2)$ so no you do not get the Stanley-Reisner ring of the circle.
Addendum: Note that cones mean somewhat different things in the simplicial and affine world. For a simplicial complex its cone is obtained by picking a new point, the apex, and connecting
with all points on the complex by using $1$-simplices, i.e., closed intervals. For the affine picture corresponding to the Stanley-Reisner ring one picks an apex and takes a line for
every point on the affine hull of the simplicial complex passing through the point and the apex but being a line continues on the other side of the apex.
Hence starting with a $1$-simplex one draws the interval between each point on the simplex and a point outside of the affine span of the $1$-simplex. In the affine picture, the
$1$-simplex is replaced by a line and one then draws all the lines through a point on that line and the apex getting the plane.
The non-homogeneous ring is obtained from the Stanley-Reisner ring by dividing out by the relation that all the variables sum up to $1$.
Thank you for your answer, Torsten. I see that the SR variety of the simplicial complex consisting of three isolated points is exactly the cone, three coordinate axes. But if you have
the 1-simplex, it corresponds to the $A^2$. In how far is this a cone? If you draw the 1-simplex as connecting $(1,0)$ and $(0,1)$, the cone is just the first and the third quadrant of
$A^2$, right? Can one get the non-homogeneous SR ring out of the homogeneous one? How? Thanks. – roger123 Dec 9 '10 at 18:03
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I think you're asking if there's a direct geometric relationship between an algebraic variety $X=V(I)$ (i.e., the zero set of an ideal $I$) and the Stanley-Reisner complex $\Delta_I$ or $\
Delta_{in(I)}$ --- in other words, does the variety look like the simplicial complex? In general, I think the answer can be quite subtle, and is probably best approached using the algebra as
an intermediary.. For example, if $\Delta$ is a simplicial sphere, then the number $d$ of its facets will correspond to the degree of $X$, but you will not necessarily be able to ``see''
directly from $X$ how those facets fit together combinatorially in $\Delta$ (e.g., if $d=20$, is $\Delta$ a decagonal bipyramid or an icosahedron?). On the other hand, you can calculate $d$
and similar invariants fairly easily from the ideal $I$ (in this case, compute the Hilbert series as a rational function in $q$, and plug $q=1$ into the numerator).
up vote For "flat deformation", here's the example I always keep in mind (very similar to Torsten's): take the hyperbola $xy=c$ and let $c\to0$. In the limit, the hyperbola degenerates into a pair of
1 down lines $x=0$ and $y=0$. This corresponds to replacing the ideal $(xy-c)$ (or $(xy-cz^2)$, if you want to think projectively) with the monomial ideal $(xy)$, which is its initial ideal under
vote the right term ordering on $\Bbbk[x,y,z]$. The general principle is that some invariants (e.g., Hilbert series) don't change when you degenerate, and some (e.g., singularities) can only get \
emph{worse}, so if you can prove that the degeneration is, say, Cohen-Macaulay then so was the original thing.
As the operation of replacing, e.g., $X_i^2\in I$ by $X_iX_i'$, it is called ``polarization'' and has similarly mild effects. See, e.g., \S3.2 of Miller and Sturmfels \emph{Combinatorial
Commutative Algebra}.
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Math Help
March 6th 2013, 12:11 AM #1
Oct 2012
If you divide a 5 digit number by a 2 digit number, how many digits will the remainder have?
It will have 2 digits but is there a practical way to solve that type of questions?
Re: division
Hey kastamonu.
There is a way and its through logarithms.
The number of digits left over after dividing m into n digit will be ln(m/n) = ln(m) - ln(n) where ln(m) is the number of digits (in fractional form) and ln(n) is the number of digits for n
(again in fractional form).
You can round the log results but that's the basic idea.
Re: division
for example 59873 by 45,
Logarithm method is not a quick way. You have to look for the log table. I need a more quick way to get the answer in 1-2 minutes.
Re: division
By definition, the remainder does not exceed the divisor, so it will have at most two digits.
Re: division
Many Thanks.
March 6th 2013, 12:29 AM #2
MHF Contributor
Sep 2012
March 6th 2013, 12:35 AM #3
Oct 2012
March 8th 2013, 12:25 PM #4
MHF Contributor
Oct 2009
March 9th 2013, 02:22 AM #5
Oct 2012 | {"url":"http://mathhelpforum.com/discrete-math/214299-division.html","timestamp":"2014-04-16T17:50:49Z","content_type":null,"content_length":"38377","record_id":"<urn:uuid:15ff1334-b864-45e0-befa-e68a1a52b313>","cc-path":"CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00524-ip-10-147-4-33.ec2.internal.warc.gz"} |
Disk seeks are evil, so let’s avoid them, pt. 3 (Deletions)
As mentioned in parts 1 and 2, having many disk seeks are bad (they slow down performance). Fractal tree data structures minimize disk seeks on ad-hoc insertions, whereas B-trees practically
guarantee that disk seeks are performed on ad-hoc insertions. As a result, fractal tree data structures can insert data up to two orders of magnitude faster than B-Trees can.
In this post, let’s examine deletions, and get an intuitive understanding for why fractal-tree data structures exhibit the same two orders of magnitude faster deletions than B-trees. In MySQL 5.1,
this advantage is really eye-popping for TokuDB v. InnoDB, because InnoDB does not use its insert buffer for deletions. I understand there is a delete buffer in 5.5, which I haven’t experimented with
B-trees exhibit the same weakness on deletions as they do on insertions: they need to have the appropriate leaf node in memory. For large tables, bringing the leaf node into memory often requires a
disk seek. Fractal tree data structures do not have this requirement.
Before going on, a clarification. In MySQL, delete statements have two steps: queries and value changes. For instance, the statement:
delete from foo where a=1;
must first query all rows where a=1 (the first step), and then proceed to remove the rows that are found (the second step). In this post, we focus on the second step. For storage engine developers,
this is the function handler::delete_row. In a future post, I will analyze the first step, tie it together with this post, and show how deletions can be fast in MySQL with TokuDB.
Back to deletions. Let’s analyze value changes. We know the contents of the row being deleted. So how can fractal tree data structures avoid an unnecessary disk seek? The answer: deletion messages
(sometimes called tombstone deletes).
Suppose we have a fractal tree data structure with the following elements inserted: (1), (3),…(999). Up until now, we shown the fractal tree as such.
- -
- - - -
1 3 5 7 9 ... 999
In reality, the elements stored are not just keys, but rather (message, key) pairs. The message may be one of two operations: insertion or deletion. We represent an insertion with the message ‘i’.
So, the fractal tree looks more like this:
- -
- - - -
(i,1) (i,3) (i,5)... (i,999)
To delete an element, for example (5), we insert a deletion message into the tree, marking it with a ‘d’. So, after deleting (5), the fractal tree data structure looks like this:
- -
- - - -
(i,1) (i,3) (i,5)... (i,999)
With this scheme, deletions are as fast as insertions, which is to say two orders of magnitude faster than insertions or deletions into a B-tree.
On queries, a message in a higher node overrides messages in lower nodes. So upon querying (5), a cursor notices that (d,5) is located higher than (i,5), and therefore the key (5) does not exist in
the fractal tree data structure. On merges, the deletion message and insertion message cancel each other out, and space is reclaimed.
So, by using deletion messages and treating deletions like insertions, fractal tree data structures can achieve the same performance boost (two orders of magnitude) over B-trees.
One Response to Disk seeks are evil, so let’s avoid them, pt. 3 (Deletions)
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Implementing Mean Curvature Flow (MCF) for an imag
Hi Roger,
There is a region in the image with only 1 colour (white in this case) and that's why Ix = Iy = 0 exactly....
"Roger Stafford" wrote in message <iv22gm$4be$1@newscl01ah.mathworks.com>...
> "Padmavathy Subramanian" wrote in message <iv1pjj$62o$1@newscl01ah.mathworks.com>...
> > Thank you so much for your code.
> > I have verified with my code and both can create a similar Gaussian kernels hx,hy,hxx,hxy,hyy.
> > Now, i can use these kernels to convolute with the input image I to generate Ix,Iy,Ixx,Ixy,Iyy.
> > What i notice is Ix And Iy contains a lot of 0s. For example at i=5,j=5, Ix(5,5)=Iy(5,5)=0. Therefore, what puzzles me is how the curvature can be computed, given the below formula:
> > k = {[(Ix^2)(Iyy)] - [2IxIyIxy] + [(Iy^2)(Ixx)]} / {[(Ix^2) + (Iy^2)]^3/2}
> > because now the denominator is 0 (Ix^2+Iy^2).
> > I really need an explanation for this. Please help!
> - - - - - - - - - -
> I have no way of advising you on that because I have not seen the data you are processing. However I find it rather strange that Ix and Iy should both be exactly zero at the same point in the
image. Are you sure about that? Perhaps they simply look zero in a display because they are small in comparison to other values being displayed along with them. Try using "format long" to display
individual values of Ix and Iy at those same points. I venture to guess they are not exactly zero.
> Also bear in mind that because you found it necessary to do filtering with gaussian kernels, the results are bound to only be approximate. You may have to experiment with choosing optimum sigma
values and best values for the range of the kernels (what I called 'm') to get the best filtering.
> In evaluating the accuracy of these gaussian kernels, I urge you to make that test I described earlier. Create an artificial image of a quadratic form:
> I(x,y) = A(x-x0)^2/2+B*(x-x0)*(y-y0)+C*(y-y0)^2/2+D*(x-x0)+E*(y-y0)+F
> for some point (x0,y0) and for various arbitrary values of A, B, C, D, E, and F, and see how close Ixx(x0,y0), Ixy(x0,y0), Iyy(x0,y0), Ix(x0,y0), and Iy(x0,y0) are respectively to A, B, C, D, and
E. I designed the code I wrote for you to do well on such a test.
> Roger Stafford | {"url":"http://www.mathworks.com/matlabcentral/newsreader/view_thread/309534","timestamp":"2014-04-24T16:48:31Z","content_type":null,"content_length":"71284","record_id":"<urn:uuid:bc45f1fa-f6a1-4046-b91d-21824c4c2af7>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00109-ip-10-147-4-33.ec2.internal.warc.gz"} |
Discrete Dynamical Networks and their Attractor Basins
Source: http://www.complexity.org.au/ci/vol06/wuensche/wuensche.html Received: 01/07/1998
Vol 6: Copyright 1998 Accepted for publication: 15/10/1998
Discrete Dynamical Networks and their Attractor Basins
Andrew Wuensche
Santa Fe Institute, 1399 Hyde Park Road,
Santa Fe, New Mexico 87501 USA,
Email: wuensch@santafe.edu
WWW: www.santafe.edu/~wuensch/
A key notion in the study of network dynamics is that state-space is connected into basins of attraction . Convergence in attractor basins correlates with order -complexity -chaos measures on
space-time patterns. A network's ``memory'', its ability to categorize, is provided by the configuration of its separate basins, trees and sub-trees. Based on computer simulations using the software
Discrete Dynamics Lab [19], this paper provides an overview of recent work describing some of the issues, methods, measures, results, applications and conjectures.
Processes consisting of concurrent networks of interacting elements which affect each other's state over time are central to a wide range of natural and artificial systems drawn from many areas of
science; from physics to biology to cognition; to social and economic organization; to computation and artificial life; to complex systems in general. The dynamics of these ``decision making''
networks depends on the connections and update logic for each element, resulting in complex feedback webs that are difficult to treat analytically. Understanding these systems depends on numerical
simulations of idealized computer models known as discrete dynamical networks.
Cellular automata (CA) are a powerful yet simple class of network, characterized by a homogeneous rule and uniform nearest neighbour connections, providing models to study processes in physical
systems such as reaction-diffusion [15], and self-organization by the emergence of coherent interacting structures[18]. By contrast, random Boolean networks (RBN) provide models for biological
systems such as neural [3] and genetic [11] networks, where connections and rules must be less constrained. In addition, the idealized networks themselves hold intrinsic interest as mathematical/
physical systems.
A key notion underlying network behavior is that state-space is organized into a number of basins of attraction, connecting states according to their transitions, and summing up the network's global
dynamics, analogous to Poincaré's ``phase portrait'' which provided powerful insights in continuous dynamics.
Figure 1: A basin of attraction (one of 15) of a random Boolean network (n=13, k=3) shown in figure 15. The basin links 604 states, of which 523 are garden-of-Eden states. The attractor period =
7, and one of the attractor states is shown in detail as a bit pattern. The direction of time is inwards from garden-of-Eden states to the attractor, then clock-wise.
The quality of dynamical behaviour of CA, from ordered to chaotic (1), is reflected by convergence in attractor basins, their characteristic in-degree , which influences the length of transients and
attractor cycles. The in-degree of a state is its number of pre-images (predecessors). Bushy subtrees with high in-degree imply high convergence and order. Sparsely branching subtrees imply low
convergence and chaos. In the case of RBN, attractor basins reveal how the network is able to hierarchically categorize state-space into separate basins, trees and sub-trees, the network's
``memory''. Changes to the network's wiring or rules change the memory categories, providing insights into learning[17, 20].
Traditionally, network dynamics has been investigated by running networks forward from many initial states to study space-time phenomenology[15], and for statistical measures on basins of attraction[
8]. More recently, exact representations of basins of attraction and sub-trees have become accessible, where algorithms directly compute the pre-images of network states, allowing the network to be
run ``backwards'' to disclose all possible historical paths[16, 17, 21]. Based on computer simulations using the software Discrete Dynamics Lab (DDLab)[19], this paper provides an overview of network
architecture, the characteristics of space-time patterns, the methods and algorithms for reconstructing basins of attraction, and related parameters, measures, results, applications and conjectures,
placing the dynamics along particular trajectories in the context of global dynamics.
Discrete dynamical networks consist of a set of elements (cells) taking inputs from each other, and changing their cell-state according to some logical function on their inputs. The connectivity is
usually sparse. The cell-state ranges over a discrete alphabet, in this paper just a binary alphabet (0 or 1) is considered. The updating is generally synchronous, though updating sequentially in a
preset order or partial order is also of interest. A partial order is a sequence of sets of cells, where updating within each set is synchronous.
1d: k=0-13. The extra asymmetric cell in even k is on the right. The wiring is shown between two time-steps.
2d: k=2-13 (k=0-1 as in 1d). Note that k=6 and k=7 define an effectively triangular grid by changing between odd and even rows. The classical von Neumann and Moore neighbourhoods are indicated.
3d: k=6-13 (k=0-5 as in 2d), shown looking up into an axonometric cage.
Figure 2: 1d, 2d and 3d neighbourhood templates defined in DDLab. In 2d and 3d, to maximize symmetry, even k does not include the central target cell.
A CA is a very regular network, sometimes described as an artificial universe with its own physics. Cells take inputs from their nearest (and next nearest) neighbours (local ``wiring'') according to
a fixed neighbourhood template, so issues of network geometry and boundary conditions are crucial. The same logical rule is applied everywhere. Figure 2 shows neighbourhood templates for 1d, 2d and
3d as applied in DDLab. An RBN relaxes these constraints, allowing arbitrary ``wiring'' and rules, as in figure 3. The number of input wires available to each cell may also vary. However, an RBN
architecture can be biased in countless of ways, described in section 4, to constrain wiring or rules. For example an RBN wiring with a constant rule, or local wiring with mixed rules can be a
generalisation of a CA.
The wiring and rules can be tailored to very specific requirements, as in models of neural networks in the cortex [3]. The wiring can be constrained within a fixed distance from each cell, which
confers meaning to network geometry and boundary conditions, whereas with completely arbitrary wiring the geometry just provides a convenient way of representing the network. A rule mix can be
constrained to sample just a few rules or rules with a particular bias, as in genetic network models[5].
Figure 3: Network wiring. top left: 1d, k=3, the wiring is shown between two time-steps. top right: 2d, k=5. bottom: 3d, k=7. In RBN, cells anywhere in the network are wired back to each position
in the ``pseudo-neighbourhood''.
Hybrid networks can be constructed by putting an RBN within a CA or vice-versa. Networks of networks can be set up, with weak interactions so they perturb each other's dynamics. The functionality for
setting up networks in these ways is present in DDLab.
The network parameters can be listed as follows:
The system size n, the number of cells in the network.
The number of input wires per cell k, or the k-mix if the connectivity is not homogeneous. The connectivity is usually sparse, i.e.
The neighbourhood template for CA, or the pseudo-neighbourhood for RBN, as in figure 2.
For RBN, how each cell is wired relative to its pseudo-neighbourhood.
The rule for CA, or the rule scheme scheme for RBN. Rules are generally defined as look-up tables.
The updating, usually synchronous. Alternatively sequential according to a defined order or partial-order.
The underlying geometry and boundary conditions, 1d, 2d, 3d, orthogonal or triangular, or some other geometry, for example a hypercube. This is essentially a function of the the neighbourhood
template and wiring scheme. The geometry for graphically representing the network may not necessarily correspond to the underlying geometry.
A CA neighborhood, or RBN pseudo-neighborhood, of size k has 14] or threshold functions.
By convention[14] the rule table is arranged in descending order of the values of neighborhoods, and the resulting bit string converts to a decimal or hexadecimal rule number. For example the k=3
rule-table for rule 30,
The rule-table for other k values are set out in a corresponding way.
For a given geometry, the behaviour space of CA depends on the size of rule-space, k=3 rule-space reduce to 88 equivalence classes[16]. The behaviour space of RBN is much greater, taking into account
possible permutations of wiring and rule schemes, but there are also RBN equivalence classes relating to these permutations[10]. In general, the number of effectively different RBN of size n cannot
exceed 9).
A state of a discrete dynamical network is the pattern of 0s and 1s at a given time-step. A trajectory is the sequence of states at successive time-steps, the systems local dynamics. Examples of 1d,
2d and 3d space-time patterns are shown in figures 4, 5 and 6. A time axis is only possible in representations of 1d or 2d systems. As well as showing cells as white(0) or black(1), an alternative
presentation shows cells in colors (or shades) according to their look-up neighbourhood (figure 4). This allows the most frequently occurring colors to be progressively filtered to show up gliders
and other space-time structures as in figure 5, which can be done interactively, on-the-fly, in DDLab for any CA. This is an alternative method to the ``computational mechanics'' approach[4].
Figure 4: Space-time patterns of a CA (n=24, k=3, rule 90). 24 time-steps from an initial state with a single central 1. Two alternative presentations are shown. Left: cells by value, white=0
black=1. Right: cells colored (or shaded) according to their look-up neighbourhood. This allows filtering, and improves the clarity of space-time patterns in 2d and 3d.
Figure 5: Space-time patterns of the k=3 rule 54 (n=150) from the same initial state showing interacting gliders , which are embedded in a complicated background. Left: cells by value. Right:
cells by neighbourhood lookup, with the background filtered.
(a) 2d 100x100 triangular grid (b) 2d 56x56 +time (c) 3d 20x20x20
Figure 6: Examples of 2d and 3d CA space patterns. (a) is an evolved time-step of a 2d CA on a k=7 triangular lattice with a reaction-diffusion rule. (b) is the 2d game-of-Life on a . (c) is a
time-step of a 3d k=7 CA with a randomly selected rule and starting from a single central 1.
A large body of literature is devoted to the study space-time patterns in CA. ``Glider'' or particle dynamics, where coherent configurations emerge and interact, provide a striking instance of
self-organization in a simple system. Such dynamics are classified as complex, in contrast to ordered or chaotic[14], a well know example being Conway's 2d ``game-of-Life''[1]. Because glider
dynamics is relatively rare in CA rule spaces, much study has focused on the few known complex rules in 1d CA. However, an unlimited source of examples are now available, found by the methods
described in sections 3.2 - 3.3.
Gliders are embedded within a uniform or periodic background or domain, and propagate at various velocities up the system's speed of light set by the neighbourhood diameter. Gliders are interpreted
as dislocation in the background or as the boundary reconciling two different backgrounds[4, 18]. Gliders may absorb or eject sub-gliders (glider-guns). Compound gliders may emerge made up of
sub-gliders re-colliding periodically. Figure 3.1 shows some examples.
Glider dynamics has been interpreted as occurring at a phase transition in rule-space between order and chaos [9], relative to the rule parameters 9] and Z[16] (see section 6.2). Input-entropy
provides a measure on space-time dynamics that allows the automatic classification of rule-space (see below).
(a)7e8696de (b)89ed7106 (c)89ed7106 (d)b51e9ce8 (e)b51e9ce8
Figure 7: Gliders, glider guns and compound gliders in k=5 1d CA . (c) is a compound glider made up of two independent gliders locked into a cycle of repeating collisions. (d) is a glider with a
period of 106 time-steps. (e) is a compound glider-gun.
Keeping track of the frequency of rule-table look-ups (the k-block frequency, or ``look-up frequency'') in a window of time-steps, provides a measure, the variance of input-entropy over time, which
is used to classify 1d CA automatically for a spectrum of ordered, complex and chaotic dynamics[22]. The method allows screening out rules that support glider dynamics and related complex rules,
giving an unlimited source for further study. The method also shows the distribution of rule classes in the rule-spaces of varying neighbourhood sizes. The classification produced seems to correspond
to our subjective view of space-time dynamics, and to global measures on the ``bushiness'' of typical sub-trees in attractor basins, characterized by the distribution of in-degree sizes in their
branching structure.
The look-up frequency can be represented by a histogram (figure 8) which distributes the total of n=system size, w=the window of time-steps defined and k=neighbourhood size. The Shannon entropy of
this frequency distribution, the ``input-entropy'' S, at time-step t, for one time-step (w=1), is given by, i at time t. In practice the measures are smoothed by being taken over a moving window of
time-steps (w=10 in figure 8).
Figure 8: Typical 1d CA Space-time patterns showing ordered, complex and chaotic dynamics (n=150, k=5, rule numbers shown in hex). Alongside each space-time pattern is a plot of the
input-entropy, where only complex dynamics (centre) exhibits high variance because glider collisions make new gliders.
Figure 8 shows typical examples of ordered, complex and chaotic dynamics in 1d CA, with input-entropy plots and a snapshot of the lookup frequency histogram alongside. In ordered dynamics the entropy
quickly settles at a low value with low or zero variance. In chaotic dynamics the entropy settles at a high value, but again with low variance. Both ordered and chaotic dynamics have low
input-entropy variance. By contrast, in complex dynamics the entropy fluctuates erratically both up and down for an extended time, because glider collisions produce new gliders, often via a temporary
zone of chaotic dynamics. Complex rules can be recognized by eye, subjectively. Input-entropy variance provides a non-subjective measure for recognizing complex rules automatically.
Figure 9: Entropy-density scatter plot. Input-entropy is plotted against the density of 1s relative to a moving window of time-steps w=10. k=5, n=150. Plots for a number of complex rules from the
automatic sample (section 3.3) are show superimposed, each of which has its own distinctive signature, with a marked vertical extent, i.e. high input-entropy variance. About 1000 time-steps are
plotted from several random initial states for each rule.
A related method of visualizing the entropy-variance is to plot input-entropy against the density of 1s relative to a moving window of time-steps. Each rule produces a characteristic cloud of points
which lie within a parabolic envelope because high entropy is most probable at medium density, low entropy at either low or high density. Each complex rule produces a plot with its own distinctive
signature, with high input-entropy variance. Chaotic rules, on the other hand, will give a flat, compact cloud at high entropy (at the top of the parabola). For ordered rules the entropy rapidly
falls off with very few data points because the system moves rapidly to an attractor.
Figure 10: top: Classifying a random sample of k=5 rules by plotting mean entropy against standard deviation of the entropy, with the frequency of rules within a middle and bottom: Equivalent
plots for samples of k=6 and 7 rules.
To distinguish ordered, complex and chaotic rules automatically, the mean input-entropy taken over a span of time-steps is plotted against the standard deviation of the input entropy. Figure 10
summarizes how random samples of k=5, 6 any 7 rules where classified by this method. For each rule, the data was gathered from 5 runs from random initial states, for 430 time-steps, discounting the
first 30 to allow the system to settle, with w=5 as the size of the moving window of time-steps.
Chaotic rules are concentrated in the top left corner ``tower", ordered rules on the left with lower entropy. Complex rules have higher standard deviation, and are spread out towards the right. There
is a fairly distinct boundary between ordered and chaotic rules, but a gradual transition from both towards the complex rules. As the standard deviation decreases glider interactions either become
more frequent, transients longer, tending towards chaos, or less frequent, transients shorter, tending towards order. The plots for k=6 and k=7 rules indicate a greater frequency of chaotic rules at
the expense of ordered and complex rules for greater k. The decrease in ordered rules is especially marked.
To check whether the expected dynamics (recognized subjectively) corresponds to the measures as plotted, the dynamics of particular rules at different positions on the plots can be easily examined in
DDLab, for example with a mouse click on the scatter plot. Preliminary scans confirm that the expected behaviour is indeed found, but further investigation is required to properly demarcate the space
between ordered, complex and chaotic rules and to estimate the proportion of different rule classes for different k.
Input entropy is a local measure on the space-time patterns of typical trajectories. The distribution of the rule samples according to these local measures may be compared with global measures on
convergence in attractor basins, G-density and the in-degree frequency, described in section 8. Preliminary results indicate a strong relationship between these global measures and the rule sample
input-entropy plots.
Figure 11: Space-time patterns for intermediate 1d architecture, from CA to RBN. n=150, k=5, 150 time-steps from a random initial state. (a) Starting off as a complex CA (rule 6c1e53a8 as in
figure 8), 4% (30/750) of available wires are randomized at 30 time-step intervals. The coherent pattern is progressively degraded. (b) A network with local wiring but mixed rules, vertical
features are evident. (c) RBN, random wiring and mixed rules, with no bias, shows maximal chaotic dynamics.
In contrast to CA, glider dynamics in general cannot occur in RBN because of their irregular architecture. Figure 11 (left) shows glider dynamics degrading as local wiring is progressively scrambled.
An alternative order-chaos notion in RBN is the balance between ``frozen'', stabilized, regions and changing regions in the space-time pattern[8]. Stable regions are characteristic of RBN with low
connectivity, 6.2) or a high proportion of ``canalizing'' inputs. In a rule's lookup table, an input wire is canalizing if a particular input (0 or 1) determines, by itself, the neighbourhood's
output. A rule's degree of canalization can be from 0 to k, for the same output; for the network it is the percentage of all inputs that are canalizing, C. An RBN's order-chaos characteristics, for
varying C, are captured by the measures illustrated in figure 12, and described below.
Figure 12: Order-chaos measures for a RBN k=5. C = the percentage of canalizing inputs in the randomly biased network. top left: frozen elements that have stabilized for 20 time-steps are shown,
0s-green, 1s red, otherwise white, for C=25% and 52%. top right: the log-log ``damage spread'' histogram for C=52%, sample size about 1000. bottom: the Derrida plot for C=0%, 25%, 52%, and 75%,
for 1 time-step,
The ``Derrida plot''[2] , is analogous to the Lyapunov exponent in continuous dynamics, and measures the divergence of trajectories based on normalized Hamming distance H, the fraction of bits that
differ between two patterns. Pairs of random states separated by 12). A curve above the main diagonal indicates divergent trajectories and chaos, below -- convergence and order. A curve tangential to
the main diagonal indicates a balance. A related measure is the distribution of ``damage spread'' resulting from a single bit change at a random position in a random state, for a sample of random
states. The size of damage is measured once it has stabilized, i.e. not changed for say 5 time-steps. A histogram (figure 12) is plotted of damage size against the frequency of sizes. Its shape
indicates order or chaos in the network, where a balance between order and chaos approximates to a power law distribution. Results by these measures for k=5, indicate a balance at 12). There are
further measures on basins of attraction as in figure 14.
These methods are applied in the context of RBN models of genetic regulatory networks[8] discussed in section 10. The conjecture is that evolution maintains genetic regulatory networks marginally on
the ordered side of the order-chaos boundary to achieve stability and adaptability in the pattern of gene expression which defines the cell type[5].
For a network size n, an example of one of its states B might be State-space is made up of all
Part of a trajectory in state-space, where C is a successor of B, and A is a pre-image of B, according to the dynamics of the network.
The state B may have other pre-images besides A, the total is the in-degree. The pre-image states may have their own pre-images or none. States without pre-images are known as garden-of-Eden
Any trajectory must sooner or later encounter a state that occurred previously - it has entered an attractor cycle. The trajectory leading to the attractor a transient. The period of the
attractor is the number of states in its cycle, which may be only just one -- a point attractor.
Take a state on the attractor, find its pre-images (excluding the pre-image on the attractor). Now find the pre-images of each pre-image, and so on, until all garden-of-Eden states are reached.
The graph of linked states is a transient tree rooted on the attractor state. Part of the transient tree is a subtree defined by its root.
Construct each transient tree (if any) from each attractor state. The complete graph is the basin of attraction. Some basins of attraction have no transient trees, just the bare ``attractor''.
Now find every attractor cycle in state-space and construct its basin of attraction. This is the basin of attraction field containing all
Figure 13: State space and basins of attraction.
The idea of basins of attraction in discrete dynamical networks is summarized in figure 13. Given invariant network architecture and the absence of noise, a discrete dynamical network is
deterministic, and follows a unique (though in general, unpredictable) trajectory from any initial state. When a state that occurred previously is revisited, which must happen in a finite
state-space, the dynamics becomes trapped in a perpetual cycle of repetitions defining the attractor (state cycle) and its period (minimum one, a stable point).
These systems are dissipative. A state may have multiple ``pre-images'' (predecessors), or none, but just one successor. The number of pre-images is the state's ``in-degree''. In-degrees greater than
one require that transient states exist outside the attractor. Tracing connections backwards to successive pre-images of transient states will reveals a tree-like topology where the ``leaves'' are
states without pre-images, known as garden-of-Eden states. Conversely, the flow in state-space is convergent. Measures of convergence are G-density, the fraction of states that are garden-of-Eden,
and the distribution of in-degrees, described in section 8. The set of transient trees rooted on the attractor is its basin of attraction (figure 1). The local dynamics connects state-space into a
number of basins, the basin of attraction field, representing the systems global dynamics (figure 15).
Figure 14: Statistical data on attractor basins for a large network; a 2d RBN 20 k=5, with fully random wiring and a fraction of canalizing inputs C=50%. The histogram shows attractor types and
the frequency of reaching each type from 12,360 random initial states, sorted by frequency. 46 different attractors types where found, their periods ranging from 4 to 102, with average transient
length from 21 to 113 time-steps. The frequency of arriving at each attractor type indicates the relative size of the basin of attraction.
Attractor basins are constructed with algorithms that directly compute the pre-images of network states[16, 17, 21]. This allows the network's dynamics, in effect, to be run backwards in time.
Backward trajectories will, as a rule, diverge. Different reverse algorithms apply to networks with different sorts of connectivity. The most computationally efficient algorithm applies to 1d
networks with local wiring, taking advantage of the regularity of connections. The wiring must be uniform, as for 1d CA, but the network may have a mix of rules. Analogous algorithms could be derived
for 2d and 3d networks, but have not been implemented. An alternative algorithm is required for RBN with their non-local connections and possibly mixed k. This algorithm also applies to CA of any
dimension or geometry, as CA are just a sub-class of RBN.
Provided 9.1), constructing an exhaustive map resulting from network dynamics, a method which rapidly becomes intractable with increasing network size and so is limited to very small systems.
However, the exhaustive method may be applied to all types of network, and also allows the attractor basins of random maps to be constructed, as described in section 9. The agreement of these three
independent methods, and other checks, give considerable confidence in the accuracy of the pre-image computations.
Some basic information on attractor basin structure can be found by statistical methods, first applied by Walker[13], as shown in figure 6. These are also implemented in DDLab and are appropriate for
large networks. Trajectories are run forward from many random initial states looking for a repeat in the network pattern to identify the range of attractor types reached. The frequency of reaching a
given attractor type indicates the relative size of the basin of attraction, and other data are extracted such as the number of basins, and the length of transients and attractor cycles.
Figure 15: The basin of attraction field of a random Boolean network (n=13, k=3). The 1 shows the arrowed basin in more detail. Right: the network's architecture, its wiring/rule scheme.
Consider a 1d CA of size n (indexed k. To find the all pre-images of a state A, let P be a ``partial pre-image'' where at least k-1 continuous bits (on the left) up to and including
If k=3 (for example), the bitstring
1. deterministic: if
2. ambiguous: if
3. forbidden: if
If forbidden (3) the partial pre-image P is rejected. If deterministic or ambiguous (1 or 2) the procedure is continued to find the next unknown bit to the right. However, in the ambiguous case (2),
both alternative partial pre-images must be continued. In practice one is assigned to a stack of partial pre-images to be continued at a later stage. As the procedure is re-applied to determine each
successive unknown bit towards the right, each incidence of ambiguous permutations will require another partial pre-image to be added to the stack. Various refinements can limit this growth.
The procedure is continued to the right to overlap the assumed start string, to check if periodic boundary conditions are satisfied; if so the the pre-image is valid. The procedure is re-applied to
each partial pre-image taken from the partial pre-image stack, starting at the first unknown cell. Each time an ambiguous permutation (2) occurs a new partial pre-image must be added to the stack,
but the stack will eventually be exhausted, at which point all the valid pre-images containing the assumed start string will have been found. The procedure is applied for k-1 bits. The reverse
algorithm is applied from left to right in DDLab, but is equally valid when applied from right to left. Examples are given in [16, 21].
A by product of the CA reverse algorithm is the probability of the next unknown bit being deterministic (section 6.1(1)). Two versions of this probability are calculated from the rule-table. Z
parameter is the greater of these values. For Z=1 it can be shown[16] that for any system size n, the maximum in-degree, k-1 may generate at most 6.1(3)). At the other extreme, for Z=0, all state
space converges on the state all-0s or all-1s in one step. For high Z, low in-degree (relative to system size n) is expected in attractor basins, growing at a slow rate with respect to n. Conversely,
for low Z, high relative in-degree is expected growing quickly with respect to n. High Z predicts low convergence and chaos, low Z predicts high convergence and order.
The k, each indexed T (0 or 1) which makes up the rule-table (section 2), and may be expressed as
The probability that the next bit is determined because of the above is given by,
The probability that the next bit is determined because of the above is given by, next bit is determined. The union of the probabilities
which simplifies to,
and may be expressed as(2) Z parameter = the greater of 16, 21].
By virtue of being a convergence parameter, Z is also an order-chaos parameter varying from 0(order) - 1(chaos). Z can be compared with Langton's[9] well known (3). c=the count of 1s a rule-table on
k inputs. Z and 16], less. Z.
Plots of G-density against both the 6.2, for the 256 k=7 totalistic rules, which reduce to 136 non-equivalent rules in 72 clusters (having equal Z). Points plotted in the top right corner of the
Figure 16: G-density against both Z for the set of k=7 totalistic rules, n=16, for
Figure 17: Computing RBN pre-images. The changing size of a typical partial pre-image stack at successive elements. n=24, k=3.
Consider an RBN of size n. Find all pre-images of a state A, k network), indexed n-1 and 0, the position of the wire connection from the jth branch of the pseudo-neighbourhood, and a rule-table
To find the all pre-images of A, let P, empty network elements, as yet unassigned to either 0 or 1. Starting with an element of A, P according to the wiring scheme
Now repeat the procedure for another element of A, say
This procedure is repeated in turn for the remaining network elements of A. If the stack size is reduced to zero at any stage A has no pre-images. The algorithm works for any ordering of elements in
A, though to minimizes the growth of the partial pre-image stack, the order should correspond to the greatest overlap of wiring schemes. The changing size of the stack at successive elements can be
displayed in DDLab, an example is shown in figure 17. When the procedure is complete, the final pre-image stack may still have empty network elements, which did not figure in any wiring scheme. These
are duplicated so that all possible configurations at empty element positions are represented. The resulting pre-image stack is the complete set of pre-images of A without duplication.
The reverse algorithm for RBN works for networks with any degree of intermediate architecture between RBN and CA, including CA of any dimension. More detailed explanations of the algorithm are given
in [17, 21].
To construct a basin of attraction containing a particular state, the network is iterated forward from the state until a repeat is found and the attractor identified. The transient tree (if it
exists) rooted on each attractor state is constructed in turn. Using one of the reverse algorithms, the pre-images of the attractor state are computed, ignoring the pre-image lying on the attractor
itself. Then the pre-images of pre-images are computed, until all ``garden-of-Eden'' states have been reached.
In a similar way, just a subtree may be constructed rooted on a state. Because a state chosen at random is very likely to be a garden-of-Eden state, it is usually necessary to run the network forward
by at least one time-step, and use the state reached as the subtree root. Running forward by more steps will reach a state deeper in the subtree so allow a larger subtree to be constructed.
For CA, a considerable speedup in computation is achieved by taking advantage of ``rotational symmetry''[16], a property of the regularity of CA and periodic boundary conditions, resulting in
equivalent subtrees and basins.
Attractor basins are portrayed as state transition graphs, vertices (nodes) connected by directed edges. States are represented by nodes, by a bit pattern in 1d or 2d, or as the decimal or hex value
of the state. In the graphic convention[16, 19], the length of edges decreases with distance away from the attractor, and the diameter of the attractor cycle approaches an upper limit with increasing
period. The direction of edges (i.e. time) is inward from garden-of-Eden states to the attractor, and then clockwise around the attractor cycle, as shown in figure 1. Typically, the vast majority of
states in a basin of attraction lie on transient trees outside the attractor, and the vast majority of these states are garden-of-Eden states.
Figure 18: The G-density plotted against system size n, for the ordered, complex and chaotic rules shown in figures 8 and 19. The the entire basin of attraction field was plotted for n=7 to 22,
and garden-of-Eden states counted. The relative G-density and rate of increase with n provides a simple measure of convergence.
Measures on attractor basins include the number of attractors, attractor periods, size of basins, characteristic length of transients and the characteristic branching within trees. The last in
particular gives a good measure of the convergence of the dynamical flow in state-space, where high convergence indicates ordered, and low convergence indicates chaotic dynamics.
The simplest measure that captures the degree of convergence is the density of garden-of-Eden states[18], G-density, counted in attractor basins or sub-trees, and the rate of increase of G-density
with n as shown in figure 18. A more comprehensive measure is the in-degree frequency distribution, plotted as a histogram. The in-degree of a state is the number of its immediate pre-images. This
can be taken on a basin of attraction field, a single basin, a subtrees, or on just part of a subtree for larger systems. Subtrees are portrayed as graphs showing trajectories merging onto the
sub-tree root state.
Examples of in-degree histograms for typical sub-trees of ordered, complex, and chaotic rules are shown in figure 19. The horizontal axis represents in-degree size, from zero (garden-of-Eden states)
upwards, the vertical axis represents the frequency of the different in-degrees. The system size n=50 for the complex and chaotic rules. For very ordered rules in-degrees become astronomical. The
ordered rule shown is only moderately ordered, however the system size was reduced to n=40 to allow easier computation.
From the preliminary data gathered so far, the profile of the in-degree histogram for different classes of rule is as follows:
Ordered rules: Very high garden-of-Eden frequency and significant frequency of high in-degrees. High convergence.
Complex rules: Approximates a power law distribution. Medium convergence.
Chaotic rules: Lower garden-of-Eden frequency compared to complex rules, and a higher frequency of low in degrees. Low convergence.
Figure 19a: Ordered dynamics. Rule 01dc3610, n=40, Z=0.5625, right: The complete sub-tree 7 levels deep, with 58153 nodes, G-density=0.931.
Figure 19b: Complex dynamics. Rule 6c1e53a8, n=50, Z=0.727, right: The sub-tree, stopped after 12 levels, with 144876 nodes, G-density=0.692.
Figure 19c: Chaotic dynamics. Rule 994a6a65, n=50, Z=0.938, right: The sub-tree, stopped after about 75 levels, with 9446 nodes, G-density=0.487.
The attractor basins of discrete dynamical networks can be put into the wider context of random graph theory. CA belong to the set of RBN which in turn belong to the set of random directed graphs
with out degree one, known as random maps. This is a mapping of the Boolean hypercube of sequences of length n (i.e. a set n), a mapping from
Figure 20:The basin of attraction field of a typical unbiased random map, n=12. The p) and size (s) of the biggest three (top row), including their percentage of state-space, are as follows: (1)
p=118 s=3204=78.2%. (2) p=20 s=599=14.6%. (3) p=32 s=194=4.74%. The field's G-density=0.37, this is a low value implying chaotic dynamics.
A random map can be constructed by assigning a successor to each state in state-space, i.e. independently assign one successor (or image)
The list of images is likely to contain repeats, and if so some other members of 20.
Random maps provide the most general context for a discrete dynamical system and are equivalent to a fully connected RBN where k=n, (the neighbourhood = the network size). This follows because each
cell in the RBN can be assigned an arbitrary output for any network state.
The ``brute force'' reverse algorithm for finding the pre-images of states in random maps can also be applied to discrete dynamical networks, RBN and CA. The method depends on first constructing an
exhaustive mapping S are found by scanning the image list; any occurrence of S in the list gives a pre-image, the state paired with S. If S does not occur in the list it has no pre-images, a
garden-of-Eden state.
Genetic regulatory networks have been thought of as discrete dynamical networks, to explain how gene expression is able to settle into a number of distinct stable patterns or cell types, despite the
fact that all eukaryotic cells in an organism carry an identical set of genes[5, 7, 11, 23]. The gene expression pattern of a cell needs to be stable but also adaptable. Section 4 described biases to
RBN to achieve such a balance, and related measures.
Cell types have been interpreted as the separate attractors or basins of attraction into which network dynamics settles from various initial states. Trajectories leading to attractors are seen as the
pathways of differentiation. The attractor basins in RBN are idealized models for the stability of cell types against mutations, and also perturbations of the current state of gene activation. Figure
21 illustrates both effects. If a particular reference state (pattern of gene activation) undergoes a 1 bit perturbation, the dynamics may return to the same subtree, the same basin, or it may be
flipped to another basin, a different cell type. In this case the basin of attraction field remains unchanged. Alternatively, the network itself my undergo a mutation (in the genotype), resulting in
an an altered basin of attraction field (the phenotype).
The examples in figure 21 are small so that the pattern at each node can be shown. Larger networks are affected in analogous ways. The consequences of a one bit mutation has a relatively smaller
effect with increasing network size. However, a particular one bit mutation may cause drastic consequences whatever the size, such as breaking an attractor cycle. The consequences of moving a
connection wire is usually greater than a one bit mutation in a rule.
Attractors classify state-space into broad categories, the network's ``content addressable'' memory in the sense of Hopfield[6]. Furthermore, state-space is categorized along transients, by the root
of each subtree forming a hierarchy of sub-categories. This notion of memory far from the equilibrium condition of attractors greatly extends the classical concept of memory by attractors alone[17,
It can be argued that in biological networks such as neural networks in the brain or networks of genes regulating the differentiation and adaptive behavior of cells, attractor basins and subtrees,
the network's memory, must be just right for effective categorization. The dynamics need to be sufficiently versatile for adaptive behavior but short of chaotic to ensure reliable behavior, and this
in turn implies a balance between order and chaos in the network.
A current research topic, known as the ``inverse problem'', is to find ways to deduce network architecture from usually incomplete data on transitions, such as a trajectory. This is significant in
genetics, to infer the genetic regulatory network (modeled as RBN) from data on successive patterns of gene expression in the developing embryo[12]. In pattern recognition and similar applications in
the area of artificial neural networks, solutions to the inverse problem would provide ``learning'' methods for RBN to make useful categories[17, 20].
Figure 21:The basin of attraction field of (a) The RBN (n=6, k=3) as defined in the table (above), and (b) the RBN following a 1 bit mutation to one of its rules. Some differences in the fields
are evident. The result or a 1 bit perturbation to a reference state of all 1s (rs) is indicated by its 1 bit mutants (m).
Important insights may be gained by considering network dynamics in the context of attractor basins. Some methods of achieving this have been presented, including parameters and measures on
particular trajectories that may be related to those on global dynamics. It is hoped that these methods may provide a basis for future research, both in theory and applications, in the many areas of
complex systems where network dynamics plays a central role.
Acknowledgments: Thanks to Cosma Shalizi for suggestions, and to the many people who have contributed to this work over the years, notably Chris Langton and Stuart Kauffman.
The notion of ``chaos'' is used here by analogy only to its meaning in chaos theory, although there are many common properties, for example sensitivity to initial conditions.
Acknowledgment and thanks to Guillaume Barreau and Phil Husbands at COGS, Univ. of Sussex, for deriving this expression.
Other versions of binary 13], and the P parameter, applied for RBN, which varies between 0.5 (chaos) - 1 (order). more,
Conway,J.H., (1982) ``What is Life?'' in Winning ways for your mathematical plays, Berlekamp,E, J.H.Conway and R.Guy, Vol.2, chap.25, Academic Press, New York.
Derrida,B., and D.Stauffer, (1986) ``Phase transitions in Two-Dimensional Kauffman Random Network Automata'', Europhys.Lett. 2, 739.
Douglas,R., and A.Wuensche, work in progress.
Hanson,J.E., and J.P.Cruchfield (1997) ``Computational Mechanics of Cellular Automata, An example'', Physica D, vol. 103, 169-189.
Harris,E.S., B.K.Sawhill, A.Wuensche, and S.Kauffman, (1997) ``Biased Eukaryotic Gene Regulation Rules Suggest Genome Behaviour is Near Edge of Chaos'', Santa Fe Institute Working Paper
Hopfield,J.J. (1982) ``Neural networks and physical systems with emergent collective computational abilities'', Proceedings of NAS 79:2554-2558.
Kauffman,S.A., (1969) ``Metabolic stability and epigenisis in randomly constructed genetic nets'', Journal of Theoretical Biology, 22, 437-467.
Kauffman,S.A., (1993) ``The Origins of Order'', Oxford University Press.
Langton,C.G., (1990) ``Computation at the Edge of Chaos, phase transitions and emergent computation'', Physica D 42, 12-37.
Myers,J.E., (1997) ``Random Boolean Networks -- Three Recent Results'', to be appear in Complexity.
Somogyi,R., and C.Sniegoski, (1996) ``Modeling the Complexity of Genetic Networks: Understanding Multigenetic and Pleiotropic Regulation'', Complexity, Vol.1/No.6,45-63.
Somogyi,R, S.Fuhrman, M.Askenazi, A.Wuensche., (1997) ``The Gene Expression Matrix'', in Proceedings of the World Congress of Non-Linear Analysis, in press.
Walker,C.C., and W.R.Ashby, (1966) ``On the temporal characteristics of behavior in certain complex systems'', Kybernetik 3, 100-108.
Wolfram,S., (1984) ``Universality and complexity in cellular automata'', Physica 10D, 1-35.
Wolfram,S., ed. (1986) ``Theory and Application of Cellular Automata'', World Scientific.
Wuensche,A., and M.J.Lesser. (1992) ``The Global Dynamics of Cellular Automata'', Santa Fe Institute Studies in the Sciences of Complexity, Addison-Wesley.
Wuensche,A., (1994) ``The Ghost in the Machine'', in Artificial Life III, ed C.G.Langton, Santa Fe Institute Studies in the Sciences of Complexity, Addison-Wesley.
Wuensche,A., (1994) ``Complexity in One-D Cellular Automata'', Santa Fe Institute Working Paper 94-04-025.
Wuensche,A., (1996) ``Discrete Dynamics Lab (DDLab)'', http://www.santafe.edu/~wuensch/ddlab.html
Wuensche,A., (1996) ``The Emergence of Memory'', in to Towards a Science of Consciousness, eds. S.R.Hameroff, A.W.Kaszniak, A.C.Scott, MIT Press.
Wuensche,A., (1997) ``Attractor Basins of Discrete Networks'', CSRP 461, Univ. of Sussex (D.Phil thesis).
Wuensche,A., (1998) ``Classifying Cellular Automata Automatically'', Santa Fe Institute Working Paper 98-02-018.
Wuensche,A., (1998) ``Genomic regulation modeled as a network with basins of attraction'', in proceedings of Pacific Symposium on Biocomputing'98, World Scientific.
EMail Contact: Complexity International Editor | {"url":"http://www.complexity.org.au/ci/vol06/wuensche/wuensche.html","timestamp":"2014-04-18T18:12:07Z","content_type":null,"content_length":"89146","record_id":"<urn:uuid:a7f55459-4239-4a81-96d6-32fd043472fd>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00000-ip-10-147-4-33.ec2.internal.warc.gz"} |
References for
References for Mikhail Egorovich Vashchenko-Zakharchenko
Version for printing
1. J W Dauben and C J Scriba, Writing the History of Mathematics: Its Historical Development (Birkhäuser, 2002).
2. L N Gratsianskaya, Mikhail Egorovich Vashchenko-Zakharchenko, in A N Bogolyubov (ed.), Mathematician-teachers from Kiev, 'Vishcha Shkola' (Kiev, 1979).
3. R A Sapsai, The work of Mikhail Egorovich Vashchenko-Zakharchenko in the field of probability theory (Russian), in A N Bogolyubov (ed.), On the history of the mathematical sciences 167 'Naukova
Dumka' (Kiev, 1984), 36-39.
4. T D Taranovskaya, The theory of determinants in the papers of Russian mathematicians of the 19th century (Russian), Istor. Metodol. Estestv. Nauk No 36 (1989), 130-138.
5. A Ziwet, Euclid as a Text-Book of Geometry, Science 4 (92) (1884), 442.
JOC/EFR April 2009
The URL of this page is: | {"url":"http://www-history.mcs.st-andrews.ac.uk/References/Vashchenko.html","timestamp":"2014-04-17T06:43:02Z","content_type":null,"content_length":"2069","record_id":"<urn:uuid:e902b03b-f501-43fd-bce7-e94198b580fa>","cc-path":"CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00191-ip-10-147-4-33.ec2.internal.warc.gz"} |
Replacing function parameters by global variables, in: J.E. Stoy (Ed
Results 11 - 20 of 61
- INFORMATION AND COMPUTATION , 1995
"... Safety analysis is an algorithm for determining if a term in an untyped lambda calculus with constants is safe, i.e., if it does not cause an error during evaluation. This ambition is also
shared by algorithms for type inference. Safety analysis and type inference are based on rather different pe ..."
Cited by 36 (6 self)
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Safety analysis is an algorithm for determining if a term in an untyped lambda calculus with constants is safe, i.e., if it does not cause an error during evaluation. This ambition is also shared by
algorithms for type inference. Safety analysis and type inference are based on rather different perspectives, however. Safety analysis is global in that it can only analyze a complete program. In
contrast, type inference is local in that it can analyze pieces of a program in isolation. In this paper we prove that safety analysis is sound , relative to both a strict and a lazy operational
semantics. We also prove that safety analysis accepts strictly more safe lambda terms than does type inference for simple types. The latter result demonstrates that global program analyses can be
more precise than local ones.
- In Proc. Fourth European Symp. Programming (ESOP’92 , 1992
"... We present an extension of a statically typed language with a special type Dynamic and explicit type tagging and checking operations (coercions). Programs in run-time typed languages are viewed
as incomplete programs that are to be completed to well-typed programs by explicitly inserting coercions i ..."
Cited by 36 (3 self)
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We present an extension of a statically typed language with a special type Dynamic and explicit type tagging and checking operations (coercions). Programs in run-time typed languages are viewed as
incomplete programs that are to be completed to well-typed programs by explicitly inserting coercions into them. Such completions are generally not unique. If the meaning of an incomplete program is
to be the meaning of any of its completions and if it is too be unambiguous it is necessary that all its completions are coherent (semantically equivalent). We characterize with an equational theory
the properties a semantics must satisfy to be coherent. Since “naive ” coercion evaluation does not satisfy all of the coherence equations we exclude certain “unsafe ” completions from consideration
that can cause avoidable type errors at run-time. Various classes of completions may be used, parameterized by whether or not coercions may only occur at data creation and data use points in a
program and whether only primitive coercions or also induced coercions. For each of these classes any term has a minimal completion that is optimal in the sense that it contains no coercions that
could be avoided by a another coercion in the same class. In particular, minimal completions contain no coercions at all whenever the program is statically typable. If only primitive type operations
are admitted we show that minimal completions can be computed in almost-linear time. If induced coercions are also allowed the minimal completion can be computed in time O(nm) where n is the size of
the program and m is the size of the value flow graph of the program, which may be of size O(n 2), but is typically rather sparse. Finally, we sketch how this explicit dynamic typing discipline can
be extended to letpolymorphism by parameterization with respect to coercions. The resulting language framework leads to a seamless integration of statically typed and dynamically typed languages by
relying on type inference for programs that have no type information and no explicit coercions whatsoever. 1
- In Proceedings of the Workshop on Principles and Practice of Constraint Programming, LNCS 874 , 1994
"... This paper contains two main parts. The first examines the set constraint calculus, discusses its history, and overviews the current state of known algorithms and related issues. Here we will
also survey the uses of set constraints, starting from early work in (imperative) program analysis, to more ..."
Cited by 35 (0 self)
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This paper contains two main parts. The first examines the set constraint calculus, discusses its history, and overviews the current state of known algorithms and related issues. Here we will also
survey the uses of set constraints, starting from early work in (imperative) program analysis, to more recent work in logic and functional programming systems. The second part describes set-based
analysis. The aim here is a declarative interpretation of what it means to approximate the meaning of a program in just one way: ignore dependencies between variables, and instead, reason about each
variable as the set of its possible runtime values. The basic approach starts with some description of the operational semantics, and then systematically replaces descriptions of environments
(mappings from program variables to values) by set environments (mappings from program variables to sets
- Information Processing Letters , 1992
"... Safety analysis is an algorithm for determining if a term in an untyped lambda calculus with constants is safe, i.e., if it does not cause an error during evaluation. We prove that safety
analysis accepts strictly more safe lambda terms than does type inference for Thatte's partial types. 1 Introduc ..."
Cited by 30 (11 self)
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Safety analysis is an algorithm for determining if a term in an untyped lambda calculus with constants is safe, i.e., if it does not cause an error during evaluation. We prove that safety analysis
accepts strictly more safe lambda terms than does type inference for Thatte's partial types. 1 Introduction We will compare two techniques for analyzing the safety of terms in an untyped lambda
calculus with constants, see figure 1. The safety we are concerned with is the absence of those run-time errors that arise from the misuse of constants. In this paper we consider just the two
constants 0 and succ. They can be misused either by applying a number to an argument, or by applying succ to an abstraction. Safety is undecidable so any analysis algorithm must reject some safe
programs. E ::= x j x:E j E 1 E 2 j 0 j succ E Figure 1: The lambda calculus. One way of achieving a safety guarantee is to perform type inference (TI), because "well-typed programs cannot go wrong".
Two examples of type ...
- ACM TRANSACTIONS ON PROGRAMMING LANGUAGES AND SYSTEMS , 1996
"... Partial-evaluation folklore has it that massaging one's source programs can make them specialize better. In Jones, Gomard, and Sestoft's recent textbook, a whole chapter is dedicated to listing
such "binding-time improvements": nonstandard use of continuationpassing style, eta-expansion, and a popul ..."
Cited by 21 (6 self)
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Partial-evaluation folklore has it that massaging one's source programs can make them specialize better. In Jones, Gomard, and Sestoft's recent textbook, a whole chapter is dedicated to listing such
"binding-time improvements": nonstandard use of continuationpassing style, eta-expansion, and a popular transformation called "The Trick". We provide a unified view of these binding-time
improvements, from a typing perspective. Just as a
, 1998
"... A useless variable is one whose value contributes nothing to the final outcome of a computation. Such variables are unlikely to occur in human-produced code, but may be introduced by various
program transformations. We would like to eliminate useless parameters from procedures and eliminate the corr ..."
Cited by 21 (1 self)
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A useless variable is one whose value contributes nothing to the final outcome of a computation. Such variables are unlikely to occur in human-produced code, but may be introduced by various program
transformations. We would like to eliminate useless parameters from procedures and eliminate the corresponding actual parameters from their call sites. This transformation is the extension to
higher-order programming of a variety of dead-code elimination optimizations that are important in compilers for first-order imperative languages. Shivers has presented such a transformation. We
reformulate the transformation and prove its correctness. We believe that this correctness proof can be a model for proofs of other analysis-based transformations. We proceed as follows: ffl We
reformulate Shivers' analysis as a set of constraints; since the constraints are conditional inclusions, they can be solved using standard algorithms. ffl We prove that any solution to the
constraints is sound: that tw...
- In Proc. 25th ACM Symposium on Principles of Programming Languages , 1998
"... The extension of Haskell with a built-in state monad combines mathematical elegance with operational efficiency: ffl Semantically, at the source language level, constructs that act on the state
are viewed as functions that pass an explicit store data structure around. ffl Operationally, at the imp ..."
Cited by 20 (2 self)
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The extension of Haskell with a built-in state monad combines mathematical elegance with operational efficiency: ffl Semantically, at the source language level, constructs that act on the state are
viewed as functions that pass an explicit store data structure around. ffl Operationally, at the implementation level, constructs that act on the state are viewed as statements whose evaluation has
the side-effect of updating the implicit global store in place. There are several unproven conjectures that the two views are consistent. Recently, we have noted that the consistency of the two views
is far from obvious: all it takes for the implementation to become unsound is one judiciously-placed beta-step in the optimization phase of the compiler. This discovery motivates the current paper in
which we formalize and show the correctness of the implementation of monadic state. For the proof, we first design a typed call-by-need language that models the intermediate language of the compiler,
- In Conference on Functional Programming Languages and Computer Architecture , 1993
"... The aggregate update problem in functional languages is concerned with detecting cases where a functional array update operation can be implemented destructively in constant time. Previous work
on this problem has assumed a fixed order of evaluation of expressions. In this paper, we devise a simple ..."
Cited by 19 (1 self)
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The aggregate update problem in functional languages is concerned with detecting cases where a functional array update operation can be implemented destructively in constant time. Previous work on
this problem has assumed a fixed order of evaluation of expressions. In this paper, we devise a simple analysis, for strict functional languages with flat aggregates, that derives a good order of
evaluation for making the updates destructive. Our work improves Hudak's work [14] on abstract reference counting, which assumes fixed order of evaluation and uses the domain of sticky reference
counts. Our abstract reference counting uses a 2-point domain. We show that for programs with no aliasing, our analysis is provably more precise than Hudak's approach (even if the fixed order of
evaluation chosen by Hudak happens to be the right order). We also show that our analysis algorithm runs in polynomial time. To the best of our knowledge, no previous work shows polynomial time
complexity. We suggest ...
, 1993
"... A binding-time analysis is correct if it always produces consistent binding-time information. Consistency prevents partial evaluators from "going wrong". A sufficient and decidable condition for
consistency, called well-annotatedness, was first presented by Gomard and Jones. In this paper we prove t ..."
Cited by 17 (6 self)
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A binding-time analysis is correct if it always produces consistent binding-time information. Consistency prevents partial evaluators from "going wrong". A sufficient and decidable condition for
consistency, called well-annotatedness, was first presented by Gomard and Jones. In this paper we prove that a weaker condition implies consistency. Our condition is decidable, subsumes the one of
Gomard and Jones, and was first studied by Schwartzbach and the present author. Our result implies the correctness of the binding-time analysis of Mogensen, and it indicates the correctness of the
core of the binding-time analyses of Bondorf and Consel. We also prove that all partial evaluators will on termination have eliminated all "eliminable"-marked parts of an input which satisfies our
condition. This generalizes a result of Gomard. Our development is for the pure -calculus with explicit binding-time annotations. 1 Introduction A partial evaluator is an implementation of Kleene's S
m n theorem....
- Lecture , 1998
"... We develop a control flow analysis algorithm for PCF based on game semantics. The analysis is closely related to Shivers' 0-CFA analysis and the algorithm is shown to be cubic. The game
semantics basis for the algorithm means that it can be naturally extended to handle strict languages and languages ..."
Cited by 16 (3 self)
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We develop a control flow analysis algorithm for PCF based on game semantics. The analysis is closely related to Shivers' 0-CFA analysis and the algorithm is shown to be cubic. The game semantics
basis for the algorithm means that it can be naturally extended to handle strict languages and languages with imperative features. These extensions are discussed in the paper. We sketch the
correctness proof for the algorithm. We also illustrate an algorithm for computing k-limited CFA. | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=309254&sort=cite&start=10","timestamp":"2014-04-21T13:47:48Z","content_type":null,"content_length":"40593","record_id":"<urn:uuid:305ec87c-5a78-4233-85c3-97fd1a6663ec>","cc-path":"CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00084-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Physical Benefits of Sports
So they say "Sports can keep you physically healthy". But, do you know why? In this section, we bring you the bolts and nuts of the physical benefits that sports brought about.
Sports can improve the 5 components of fitness, namely: Strength, speed, skill, stamina, and suppleness. Sports can burn calories. Sports can affect your appetite. The above topics will be covered
in-depth in the following passages. To get started, just click on the links below.
Improves Fitness
'Fitness' can be defined as the ability to carry out effectively and efficiently some particular physical or activity. 'Getting fitter' means advancing to a higher standard of fitness and being 'very
fit' means possessing a standard of fitness better than most people. A sound standard of fitness is conductive to a productive life and enhances the possibility of reaching your full potential. You
do not have to be trained to the level of a champion athlete to appreciate the benefits of fitness. To students, like all of us, fitness could mean a energetic state that helps us to archive. More
technically speaking, 'fitness ' consists of five components, namely: stamina, strength, speed, skill and suppleness.
Calories Burning
Whether people gain or lose weight depends on the balance between energy consumed in food and the energy used in living. This is measured in 'calories'. A calorie is defined as a unit of work or
energy equating to the amount of heat required to raise the temperature of one gram of water one degree centigrade. When the body is in calorie balance the energy intake and output are equal and the
body weight remains constant. Eat more calories than you use and weight will be gained; use more than you eat and weight will be lost. Being active is not only good for the heart and circulation, it
can use up many calories, which results in weight being controlled.
People vary in the amount of energy they use when doing similar activities. The figures on the table on the next page give a general outline as to the amount of energy used daily and may help you
calculate a personal daily diet/exercise programme.
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Learning algebra is a little like learning another language. In fact, algebra is a simple language, used to create mathematical models of real-world situations and to handle problems that we can't
solve using just arithmetic. Rather than using words, algebra uses symbols to make statements about things. In algebra, we often use letters to represent numbers.
Since algebra uses the same symbols as arithmetic for adding, subtracting, multiplying and dividing, you're already familiar with the basic vocabulary.
In this lesson, you'll learn some important new vocabulary words, and you'll see how to translate from plain English to the "language" of algebra.
The first step in learning to "speak algebra" is learning the definitions of the most commonly used words.
Algebraic Expressions | Variables | Coefficients | Constants | Real Numbers | Rational Numbers | Irrational Numbers | Translating Words into Expressions
Algebraic Expressions
An algebraic expression is one or more algebraic terms in a phrase. It can include variables, constants, and operating symbols, such as plus and minus signs. It's only a phrase, not the whole
sentence, so it doesn't include an equal sign.
Algebraic expression:
3x2 + 2y + 7xy + 5
In an algebraic expression, terms are the elements separated by the plus or minus signs. This example has four terms, 3x2, 2y, 7xy, and 5. Terms may consist of variables and coefficients, or
In algebraic expressions, letters represent variables. These letters are actually numbers in disguise. In this expression, the variables are x and y. We call these letters "variables" because the
numbers they represent can vary—that is, we can substitute one or more numbers for the letters in the expression.
Coefficients are the number part of the terms with variables. In 3x2 + 2y + 7xy + 5, the coefficient of the first term is 3. The coefficient of the second term is 2, and the coefficient of the third
term is 7.
If a term consists of only variables, its coefficient is 1.
Constants are the terms in the algebraic expression that contain only numbers. That is, they're the terms without variables. We call them constants because their value never changes, since there are
no variables in the term that can change its value. In the expression 7x2 + 3xy + 8 the constant term is "8."
Real Numbers
In algebra, we work with the set of real numbers, which we can model using a number line.
Real numbers describe real-world quantities such as amounts, distances, age, temperature, and so on. A real number can be an integer, a fraction, or a decimal. They can also be either rational or
irrational. Numbers that are not "real" are called imaginary. Imaginary numbers are used by mathematicians to describe numbers that cannot be found on the number line. They are a more complex subject
than we will work with here.
Rational Numbers
We call the set of real integers and fractions "rational numbers." Rational comes from the word "ratio" because a rational number can always be written as the ratio, or quotient, of two integers.
Examples of rational numbers
The fraction ½ is the ratio of 1 to 2.
Since three can be expressed as three over one, or the ratio of 3 to one, it is also a rational number.
The number "0.57" is also a rational number, as it can be written as a fraction.
Irrational Numbers
Some real numbers can't be expressed as a quotient of two integers. We call these numbers "irrational numbers". The decimal form of an irrational number is a non-repeating and non-terminating decimal
number. For example, you are probably familiar with the number called "pi". This irrational number is so important that we give it a name and a special symbol!
Pi cannot be written as a quotient of two integers, and its decimal form goes on forever and never repeats.
Translating Words into Algebra Language
Here are some statements in English. Just below each statement is its translation in algebra.
the sum of three times a number and eight
3x + 8
The words "the sum of" tell us we need a plus sign because we're going to add three times a number to eight. The words "three times" tell us the first term is a number multiplied by three.
In this expression, we don't need a multiplication sign or parenthesis. Phrases like "a number" or "the number" tell us our expression has an unknown quantity, called a variable. In algebra, we use
letters to represent variables.
the product of a number and the same number less 3
x(x – 3)
The words "the product of" tell us we're going to multiply a number times the number less 3. In this case, we'll use parentheses to represent the multiplication. The words "less 3" tell us to
subtract three from the unknown number.
a number divided by the same number less five
The words "divided by" tell us we're going to divide a number by the difference of the number and 5. In this case, we'll use a fraction to represent the division. The words "less 5" tell us we need a
minus sign because we're going to subtract five.
3/10/2011 تم النشر بواسطة هشام علام.
قد يهمك أيضًا
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I was posed the question
With the preposition of y > z < x > 1
Solve for x where zx + x = y
Naturally, as maths would in simple terms dictate the answer:
x = y/z+1
Computers on the other hand would go through the convoluted length of controlling hyperbolic paraboloids as:
[memory] int / ( [+increment] [(pointer to mem)] )
x = [1 / z] * y / (1 + [(1/z)])
I only question it?
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Finding Volume using spherical coordinates
March 26th 2010, 03:39 PM #1
Nov 2009
Finding Volume using spherical coordinates
Question: Find the spherical coordinate limits for the integral that calculates the volume of the solid between the sphere $\rho=4 cos\phi$ and the hemisphere $\rho=6, z\geq0$. Then evaluate the
I found the integral with the correct limits for the solid in spherical coordinates as the following: $V=\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi d\theta$
I would like to check my volume, the answer which i got is $\frac{26\pi}{3}$.
Last edited by Belowzero78; March 26th 2010 at 05:11 PM.
As long as all your bounds are correct for the figure, your answer is off. I obtained 400*pi/3 for that integral as is.
Alright, do you mind to show a few steps?
If you show me your steps, it will be quicker for me to find your error.
Question: Find the spherical coordinate limits for the integral that calculates the volume of the solid between the sphere $\rho=4 cos\phi$ and the hemisphere $\rho=6, z\geq0$. Then evaluate the
I found the integral with the correct limits for the solid in spherical coordinates as the following: $V=\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi d\theta$
I would like to check my volume, the answer which i got is $\frac{26\pi}{3}$.
$2\pi\int_{0}^{\pi/2}\int_{4cos\phi}^{6} \rho^2 sin\phi\,d\rho d\phi$
= $2\pi\int_{0}^{\pi/2} \frac{\rho^3}{3}\sin\phi \bigg|^{6}_{4\cos\phi} d\phi$
= $2\pi\int_{0}^{\pi/2} 72\sin\phi -\frac{64\cos^3\phi\sin\phi}{3}\ d\phi$
= $2\pi( -72\cos\phi +\frac{64\cos^4\phi}{12}) \bigg|^\frac{\pi}{2}_{0}$
= $2\pi(72-\frac{16}{3}) = \frac{400\pi}{3}$
March 26th 2010, 06:04 PM #2
MHF Contributor
Mar 2010
March 26th 2010, 06:11 PM #3
Nov 2009
March 26th 2010, 06:13 PM #4
MHF Contributor
Mar 2010
March 26th 2010, 06:35 PM #5 | {"url":"http://mathhelpforum.com/calculus/135847-finding-volume-using-spherical-coordinates.html","timestamp":"2014-04-21T06:04:56Z","content_type":null,"content_length":"44098","record_id":"<urn:uuid:5bbb53df-20ca-4487-aa56-ca17afc849ba>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00121-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Topic: using lab test data in simulink to simulate a variable load
Replies: 1 Last Post: Dec 29, 2012 5:49 PM
Messages: [ Previous | Next ]
isaac Re: using lab test data in simulink to simulate a variable load
Posted: Dec 29, 2012 5:49 PM
Posts: 1
Registered: "James Kende" wrote in message <g5o2ie$422$1@fred.mathworks.com>...
12/29/12 > Hello,
> i am trying to use data obtained from a test cycle.
> This data is provided in an excel sheet and contains power
> usage..ie how much power is required from the battery to
> push a load.
> To simplify things..i am trying to simulate a load using
> real data...so the load is variable..(a programmable load).
> My circuit is simple..consists of a battery (LIOH), a load
> ( my data) and measuring equipment.
> The purpose of the experiment is to test the dynamics of
> the battery.
> Please help and thanks in advance.
You can use a current source as the load with the s input coming from a Timer block. In the timer block you can input both the time and the load arrays. What I did was, imported the load
data. Divided the load (W, KW etc) by the voltage( which i am assuming is kept constant). Then, you get the current with respect to time. Put the current and the time in two arrays and
use the variable as the input in the timer block.
data = load('load.txt'); %your load data.
X = data(1,:); % Time
Y = data(2,:); % Load
current = Y./120; %Dividing by 120 V
Now, use the variables 'X' and 'current' in the timer block. Hook the timer block to the current source and you should be good.
Date Subject Author
7/17/08 using lab test data in simulink to simulate a variable load James Kende
12/29/12 Re: using lab test data in simulink to simulate a variable load isaac | {"url":"http://mathforum.org/kb/message.jspa?messageID=7945082","timestamp":"2014-04-16T19:07:03Z","content_type":null,"content_length":"18446","record_id":"<urn:uuid:87e7552a-f039-4927-b1a5-eab67b82de79>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00141-ip-10-147-4-33.ec2.internal.warc.gz"} |
Large Bore Orifice Flowmeter for Liquids
│ │ For pipe diameter > 5 cm. │
│Large Diameter Orifice Flowmeter Calculation for Liquid Flow│Compute flowrate, orifice diameter, or differential pressure. │
│ │ Equations: ISO 5167 │
│ Other Flowmeter Calculations using standard methodologies: │
│ Orifice for liquids (D<5cm) │
│ Orifice for gases (D<5 cm) Orifice for gases (D>5cm) │
│ Nozzle for liquids Venturi for liquids │
│ Simpler orifice calculation (not as accurate but won't give "parameter out of range" messages): Bernoulli page │
│ To LMNO Engineering home page Unit Conversions Page Trouble viewing or printing? │
Register to enable "Calculate" button.
Types of Pressure Taps for Orifices:
Topics: Introduction Equations Discharge Coefficients Error Messages References
Orifice flowmeters are used to determine a liquid or gas flowrate by measuring the differential pressure (P[1] - P[2]) across the orifice plate. They are generally less expensive to install and
manufacture than the other commonly used differential pressure flowmeters; however, nozzle and venturi flow meters have the advantage of lower pressure drops. Equations for orifice meters have the
advantage of no Reynolds Number upper limit for validity.
An orifice flowmeter is typically installed between flanges connecting two pipe sections (flanges are not shown in the above drawings). The three standard pressure tapping arrangements are shown in
the drawings; the location of the pressure taps affects the discharge coefficient somewhat. Flange pressure taps penetrate the flange and are at a standard distance of 1 inch (2.54 cm) from either
side of the orifice. For corner taps or D and D/2 taps, the pressure tap locations are as shown.
Orifices are typically less than 0.05D thick. For exact geometry and specifications for orifices, see ISO (1991) or ASME (1971). The ASME and ISO have been working on guidelines for orifices since
the early 1900s. The organizations have the most confidence in orifice accuracy when the Reynolds number exceeds 10^5, though Reynolds numbers as low as 4x10^3 are valid for certain d/D ratios as
discussed below. The calculation above is for liquids. Orifice gas flow calculations (D<5 cm, D>5 cm) have an additional factor called expansibility.
The calculations on this page are for orifices carrying a liquid as described in ISO (1991) and ASME (1971). The ISO reference has a more up-to-date discussion of orifices than the ASME reference, so
the ISO equations are used in our calculations.
k = Equivalent Roughness of the pipe material [L].
w is the static pressure loss occurring from a distance of approximately D upstream of the orifice to a distance of approximately 6D downstream of the orifice. It is not the same as differential
pressure. Differential pressure is measured at the exact locations specified in ISO (1991) (shown in the above figures). K[m] is computed to allow you to design pipe systems with orifices and
incorporate their head loss. Head loss is computed as h=K[m]V^2/2g where V is the pipe velocity.
Discharge Coefficients (ISO, 1998)
Equation and applicability:
Corner Pressure Taps: L[1] = L'[2] = 0
D and D/2 Pressure Taps: L[1] = 1 and L'[2] = 0.47
Flange Pressure Taps: L[1] = L'[2] = 0.0254/D where D is in meters
All types of pressure taps: d >= 1.25 cm, 5 cm <= D <= 1 m, 0.1 <= d/D <= 0.75
Corner Pressure Taps or D and D/2 Pressure Taps:
Re[D] >= 4000 for 0.1 <= d/D <= 0.5 and Re[D] >= 16,000(d/D )^2 for d/D>0.5
Flange Pressure Taps: Re[D] >= 4000 and Re[D] >= 170,000 D (d/D )^2 where D is in meters
In addition, ISO recommends that in general k/D <= 3.8x10^-4 for Corner Taps and k/D <= 10^-3 for Flange or D and D/2 pressure taps. k is the pipe roughness.
Error Messages given by calculation
"All inputs must be positve". This is an initial check of user input.
"d, D, d/D, or ReD out of range". The equation for discharge coefficient, C, is only valid for certain ranges of d, D, d/D, and ReD as shown in the Discharge Coefficients section above. This message
could be generated during the program's initial check for valid input. If the input is valid, the message will be generated during computations if the program determines that a calculated value will
be out of range of the validity of the equations. d is orifice diameter, D is pipe diamter, and ReD is the Reynolds number based on D.
· Try the simpler orifice calculation on our Bernoulli page if your parameters are out of range. It is not as accurate, but won't give "parameter out of range" error messages.
American Society of Mechanical Engineers (ASME). 1971. Fluid meters: Their theory and application. Edited by H. S. Bean. 6ed. Report of ASME Research Committee on Fluid Meters.
International Organization of Standards (ISO 5167-1). 1991. Measurement of fluid flow by means of pressure differential devices, Part 1: Orifice plates, nozzles, and Venturi tubes inserted in
circular cross-section conduits running full. Reference number: ISO 5167-1:1991(E).
International Organization of Standards (ISO 5167-1) Amendment 1. 1998. Measurement of fluid flow by means of pressure differential devices, Part 1: Orifice plates, nozzles, and Venturi tubes
inserted in circular cross-section conduits running full. Reference number: ISO 5167-1:1991/Amd.1:1998(E).
© 1999-2002 LMNO Engineering, Research, and Software, Ltd. (All Rights Reserved)
LMNO Engineering, Research, and Software, Ltd.
7860 Angel Ridge Rd. Athens, Ohio 45701 USA (740) 592-1890
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Summary: Math 140
Introductory Statistics
Professor Bernardo Ábrego
Lecture 26
Sections 8.3
8.3 A Confidence Interval for the
Difference of two Proportions
A very common and important situation
involves taking two samples independently
from two different populations with the goal of
estimating the size of the difference between
the proportion of successes in one population
and the proportion of successes in the other.
A recent poll of 29,700 U.S. households
found that 63% owned a pet. The percentage
in 1994 was 56%. [Source: American Pet Products
Manufacturers Association, www.appma.org.]
The two populations are the households in
the United States in 1994 and the households | {"url":"http://www.osti.gov/eprints/topicpages/documents/record/539/0207416.html","timestamp":"2014-04-21T07:07:51Z","content_type":null,"content_length":"7746","record_id":"<urn:uuid:89639723-ee96-46c2-9660-6c06bec939de>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00476-ip-10-147-4-33.ec2.internal.warc.gz"} |
Proceedings of the American Mathematical Society
ISSN 1088-6826(online) ISSN 0002-9939(print)
Duals and topological center of a class of matrix algebras with applications
Author: G. H. Esslamzadeh
Journal: Proc. Amer. Math. Soc. 128 (2000), 3493-3503
MSC (2000): Primary 43A20, 46H05
Published electronically: May 18, 2000
MathSciNet review: 1694860
Full-text PDF Free Access
Abstract | References | Similar Articles | Additional Information
Abstract: We characterize the topological center of a class of matrix algebras, which are called
Similar Articles
Retrieve articles in Proceedings of the American Mathematical Society with MSC (2000): 43A20, 46H05
Retrieve articles in all journals with MSC (2000): 43A20, 46H05
Additional Information
G. H. Esslamzadeh
Affiliation: Faculty of Mathematics and Computer Science, Tehran Polytechnic University, 424 Hafez Avenue, 15914 Tehran, Iran
Email: hesslam@cic.aku.ac.ir
DOI: http://dx.doi.org/10.1090/S0002-9939-00-05521-0
PII: S 0002-9939(00)05521-0
Received by editor(s): January 22, 1999
Published electronically: May 18, 2000
Additional Notes: This work was supported by a scholarship from MCHE, Iran, and also partially by the Department of Mathematical Sciences, University of Alberta. The author thanks both of these
agencies for their kind support.
Communicated by: Dale Alspach
Article copyright: © Copyright 2000 American Mathematical Society | {"url":"http://www.ams.org/journals/proc/2000-128-12/S0002-9939-00-05521-0/","timestamp":"2014-04-16T19:34:26Z","content_type":null,"content_length":"38650","record_id":"<urn:uuid:5b849a5e-89c4-410d-82f4-c5f6e6a5dac4>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00492-ip-10-147-4-33.ec2.internal.warc.gz"} |
Set Operation: Union - Problem 2
Solving a union of linear inequalities. So we have two statements here, the first thing we need to do is solve each of these inequalities as we would any to other inequalities. So for this one here
we need to add one to both sides -3x is greater than 6 divide by -3, remember when we divide by a negative we have to flip our sign we end up with x is less than -2.
Going to this one over here, same thing we are solving for x add 3 to both sides 4x is less than 16 divide by 4, x is less than 4. Whenever I am solving a linear inequality union or intersection, I
always make a number line. We have our -2 and we have our 4. X is less than -1 sorry x is less than -2 so this is going to be an open circle. And we are going down. Always feel when I’m doing two
things on the same number line I always bring them up so I can just sort of see what’s each one is doing instead of having them on top of each other.
This one we have x is less than 4 again we have a open circle and this is going to be shaded down as well. For this we are dealing with the union this union carries down union is where at least one
element is represented. So looking at this number line if we are looking at numbers less than -2 , we have both elements we have both inequalities satisfying that so that’s an union.
Looking in the middle between -2 and 4 just this one any inequalities represented there? That’s perfectly fine we only need one and lastly over here there’s nothing so this part isn’t in the union
but everything else is. So our answer then is going to be everything less than 4 not including 4 because we don’t have an equal to so soft bracket soft bracket.
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Need some ideas on how to do a raffle dr
Here's a pretty simple way to go about it. I'll give you some basic thought process since you know what you're doing :-)
So after you fill the vector of type Person with a bunch of "Persons" you're going to need to determine a winner. What you could do to simulate the "raffle" of the tickets is pick a random index
within the bounds of the vector(don't want any run-time errors). I suggest using a for loop. If you don't want to do it that way then just select a random index. I assume you know of rand in C++?
What do you mean when you say optimal? Can you be more specific?
Last edited on
Yes By optimal just meaning as a good way of going about it without hundreds of lines of code -- I know it can be done I am just not good enough to figure it out lol.
The way I am thinking about doing it is keeping a running total of how many tickets have been purchased and then going back and looping through and finding a winner.
For example, the way I am thinking about doing it.. if there were 5 entrance each bought 500 tickets.
I take a random number and get say 2250. My loop would go through each element, starting at the bottom/first entrance, seeing he has 500 and setting a temp running total number to 500. Loop would hit
the second guy and see he also had 500 making the temp running total 1000. Third guy 500 temp running total # 1500. Fourth guy 500 temp running # 2000. And finally it would hit the winner since the
temp running (2500) total went over the random number (2250) so I know that person had won.
If that makes any sense, that's how I think I want to go about it, but I'm not sure if there is an easier way... it seems like a complicated way of going about it.
Perhaps there's an easier way of doing it by summing the struct members inside of a vector?
edit: hmm a little more searching and it looks like std::accumulate is going to do mostly what I want and save me trouble accumulating tickets, but I'm still not sure the best way to pick the winner,
and more specifically, match the random winning ticket with the person.
Last edited on
Do you really need to keep track of how many elements were entered using std::accumulate ;)
try [vectornamehere].size();
You're good enough to do it. Programming isn't hard it's just a lot of thinking, "What can I do better","What happens if I do this". If you have a bug then it's a whole different story! :P
Last edited on
Have you tried std::random_shuffle?
I just looked at std::random_shuffle and while that would work if everyone bought the same amount of tickets I don't think it would work with variable amounts of tickets. It looks like it would give
the person who bought 1 ticket the same odds as the person who bought 500 tickets.
all the tickets have different numbers yes or can one person have ticket 1 and another person have ticket 1 also? From the looks of it you have it so each person has only one ticket so you would have
to type in the same name multiple times with different numbers right now.
Last edited on
I'm not sure I understand the question completely so I'll give you an example of how I have it set up now:
I have 3 participants in the raffle. The user can input as many names as they want and they are all free to purchase as many tickets as they want, but for this example, they are as follows:
Craig who has bought 100 tickets.
Joe who has bought 1 tickets.
John who has bought 2 tickets.
myVector[0].name is Craig, myVector[0].tickets is 100;
myVector[1].name is Joe, myVector[1].tickets is 1;
myVector[2].name is John, myVector[2].ticket is 2;
It would appear that if I did std::random_shuffle it would give everyone an equal opportunity to win, when clearly Craig should win most of the time.
I would like to be able to pick a random number that is 1 to Max tickets bought and match it to it's ticket holder. In the previous, anything 1-100 would hit Craig as a winner, and 101 would be Joe,
and 102&103 would be John.
Last edited on
My line 8 needs to be fixed you need to add the previous tickets before that one and figure out who has what ticket numbers after you figure out that you can see who won.
Yea that helps thank you. So that part of code will get me a random ticket number out of the maximum possible tickets. But like you said that still doesn't get me where I need to be. I'm still lost
how I would then take that random number and find it's owner.
Awesome. Just glancing over it that looks like it will do exactly what I want. Bugs/errors are no problem I can sort them... heck I was expecting pseudo code so real code is great.
Tbh I didn't even read your second post but I did pretty much exactly what you said there haha.
Lol yea. I like how you added the 'count'. That'll be super hand if I ever want to implement multiple winners, which I do. Thanks!
haha no problem. Glad I could be of help.
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Mass-Spring Oscillator Time-Domain Solution
Next | Prev | Up | Top | JOS Index | JOS Pubs | JOS Home | Search
Consider now the mass-spring oscillator:
Electrical equivalent-circuit:
Newton's second law of motion:
Hooke's law for ideal springs:
Newton's third law of motion:
We have thus derived a second-order differential equation governing the motion of the mass and spring. (Note that
Taking the Laplace transform of both sides of this differential equation gives
Solving for
denoting the modulus and angle of the pole residue
Next | Prev | Up | Top | JOS Index | JOS Pubs | JOS Home | Search Download Laplace.pdf
Download Laplace_2up.pdf
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STAT 291 Statistics for the Mathematical Sciences I
STAT 291 Statistics for the Mathematical Sciences I (3)
Topics in probability and statistics: descriptive methods, conditional and unconditional probability, discrete and continuous distributions, one-sample estimation and hypothesis testing. Meets
the General Education Mathematical Sciences requirement. (Offered fall semester only.)
MATH 108 Intuitive Calculus with Applications or MATH 111 Calculus I with a C- or better
1. Comfort with quantitative topics.
2. Familiarity with summation notation and usage. This notation is used in both descriptive analysis and when working with discrete distributions.
3. Understanding of elementary proofs.
4. Ability to differentiate and integrate functions of one variable so that probabilities and moments can be considered for practical continuous distributions.
Required Course Materials:
J. Devore, Probability and Statistics for Engineering and the Sciences, 8^th edition, Cengage, 2012 (ISBN: 9780538733526)
L. Marlin Eby, Ph.D., Professor of Mathematics and Statistics
1. Students majoring in Mathematics, Mathematics-with-Certification, Physics, Physics-with-Certification.
2. Students who select this course from a required menu majors in Computer and Information Science with Computer Science concentration, Economics (B.A.),
Education-with-Certification (4-8) with Mathematics concentration, Education-with-Certification (4-8) with Science and Mathematics concentration, Nutrition Science.
3. Students who select this course, a more mathematical introductory statistics course than STAT 269, to meet the General Education Mathematical Sciences requirement.
1. To become familiar with both descriptive and inferential analyses.
2. To use probability in applied models and as the bridge between descriptive and inferential analysis.
3. To intuitively understand each concept
4. To understand when possible and appropriate, the rigor of a mathematical proof.
5. To integrate topics by identifying commonalties.
6. To understand the limitations of each analysis through consideration of assumptions.
7. To express general concepts in terms of the application.
8. To communicate results, clearly and completely, in a manner appropriate to nonquantitative audiences.
9. To be introduced to the computer's capabilities in solving practical problems, using the computer for analysis only after understanding how to perform the analysis manually.
General Education Objectives for the Mathematical Sciences Courses:
1. To identify methods and assumptions of the mathematical sciences.
2. To understand at least one of the three mathematical sciences of computing, mathematics, and statistics
from a liberal arts perspective.
3. To think logically, analytically, and abstractly through engagement in quantitative problem-solving
1. Descriptive Statistics: pictorial and tabular methods and measures of location and variability, and position.
2. Probability: axioms and properties, counting techniques, conditional probability, and independence.
3. Discrete Distributions: probability and moment calculations in general; binomial and Poisson distributions.
4. Continuous Distributions: probability and moment calculations in general; normal, gamma (including exponential and chi-squared), and beta distributions.
5. Sampling Distributions: Central Limit Theorem.
6. One-Sample Interval Estimation: properties and assumptions; confidence intervals for a mean, proportion, and variance (standard deviation); sample size determination.
7. One-Sample Hypothesis Testing: properties and assumptions; tests on a mean, proportion, and variance (standard deviation); power calculation; relationship to interval estimation; sample size
Revised: October 2013; August 2011 (textbook) | {"url":"http://www.messiah.edu/departments/mathsci/courses/syllabi/STAT291.html","timestamp":"2014-04-16T15:59:13Z","content_type":null,"content_length":"7143","record_id":"<urn:uuid:31c9e92f-45a4-4c6a-9cc1-a013118a9672>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00394-ip-10-147-4-33.ec2.internal.warc.gz"} |
Heinrich Martin Weber
Born: 5 May 1842 in Heidelberg, Germany
Died: 17 May 1913 in Strasbourg, Germany (now France)
Click the picture above
to see three larger pictures
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Heinrich Weber was born in Heidelberg, the son of G Weber who was an historian. He entered the University of Heidelberg in 1860 but, as was the common practice of German students at this time, he
spent part of his time studying at a different university. For Weber it was the University of Leipzig that he moved to in the middle of his studies, spending a year there before returning to
Heidelberg to complete his education. He was awarded a doctorate from the University of Heidelberg in 1863.
In order to become a university teacher, Weber needed to write a further thesis, his habilitation thesis. He went to Königsberg where he studied under Franz Neumann and Friedrich Richelot, who had
been a student of Jacobi. Although Jacobi had died over ten years before Weber began his studies at Königsberg, his influence was still strongly felt and it would not be unreasonable to say that
Weber, through his teachers at Königsberg, was strongly influenced by Jacobi's style of mathematics. There were other students at Königsberg at this time who would become important in the development
of mathematics, in particular Wangerin studied for his doctorate around the same time as Weber worked for his habilitation.
In 1866 Weber's habilitation thesis was accepted and he became a privatdozent at Heidelberg in that year. Three years later he was appointed as extraordinary professor at Heidelberg. Over the next
twenty-five years, Weber taught at a number of different institutions. He taught in Zurich at the Eidgenössische Polytechnikum, at the University of Königsberg, and at the Technische Hochschule in
Charlottenburg. His final post was in Strasbourg where he was appointed in 1895.
The city of Strasbourg was German at this time (and called Strassburg) and it had been since it was annexed by Germany during the Franco-Prussian War of 1870-71. Weber remained in Strasbourg for the
rest of his life and during this time it remained a German city; only at the end of the World War I in 1918 did the city revert to France
Weber's main work was in algebra, number theory, analysis and applications of analysis to mathematical physics. This seems a contradiction in terms, for we have now almost said that Weber's main work
spans the whole spectrum of mathematics. In fact this is not far from the truth for Weber work was characterised by its breadth across a wide range of topics.
To a certain extent this breadth can be attributed to the various influences on Weber from colleagues around him. The applications to mathematical physics certainly grew from working with Franz
Neumann in Königsberg. But in Königsberg there was also the Jacobi influence, particularly coming through one of his other teacher Friedrich Richelot, which saw Weber doing important work on
algebraic functions.
Perhaps Weber is best known for his outstanding text Lehrbuch der Algebra published in 1895 and it is his work in algebra and number theory for which he is best known. If he was influenced by his
colleagues to work in different areas of mathematics then it is a very fair question to ask where the influence came from which prompted Weber to work on algebra and number theory. The answer must be
Dedekind. He wrote an important paper with Dedekind in 1882 in which they examined algebraic functions from an algebraic rather than analytic point of view.
Weber's Lehrbuch der Algebra is an outstanding work but, although he tried hard to connect the various algebraic theories, even fundamental concepts such as a field and a group are only seen as tools
and not properly developed as theories in their own right. It was, however, a remarkable book which was effective over many years as a teaching tool. In fact [1]:-
Weber was an enthusiastic and inspiring teacher who took great interest in educational questions.
Article by: J J O'Connor and E F Robertson
Click on this link to see a list of the Glossary entries for this page
List of References (4 books/articles)
Mathematicians born in the same country
Cross-references in MacTutor
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History Topics Societies, honours, etc. Famous curves
Time lines Birthplace maps Chronology Search Form
Glossary index Quotations index Poster index
Mathematicians of the day Anniversaries for the year
JOC/EFR © May 2000 School of Mathematics and Statistics
Copyright information University of St Andrews, Scotland
The URL of this page is: | {"url":"http://www-history.mcs.st-and.ac.uk/history/Biographies/Weber_Heinrich.html","timestamp":"2014-04-20T18:24:36Z","content_type":null,"content_length":"13753","record_id":"<urn:uuid:6d64d523-720f-498b-be02-371619ad961a>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00102-ip-10-147-4-33.ec2.internal.warc.gz"} |
[ODE] Power, torques and feedback
Jean-baptiste MOURET jean-baptiste.mouret at lip6.fr
Mon Sep 11 09:32:18 MST 2006
This seems to be a recurring question but I cannot find any definitive
answer; as a lot of ODE users, I have to compute the energy used by my
simulated robots to move and I need some help to set up this
I am trying to use the provided feedback functions as follow:
- I use angular motors by setting the dParamVel
- Since power is Torque dot AngularVelocity, I get the torque and the
angular velocities for the two connected bodies.
This gives this code :
dJointGetFeedback (_ball);
maths::Vectorf t1(_feedback.f1[0], _feedback.f1[1], _feedback.f1[2]);
const dReal* r1 = dBodyGetAngularVel(_o1.get_body());
maths::Vectorf v1(r1[0], r1[1], r1[2]);
float e1 = t1 * v1; // * is the dot product
maths::Vectorf t2(_feedback.f2[0], _feedback.f2[1], _feedback.f2[2]);
const dReal* r2 = dBodyGetAngularVel(_o2.get_body());
maths::Vectorf v2(r2[0], r2[1], r2[2]);
float e2 = t2 * v2; // * is the dot product
_power = e1 + e2;
=> I am really not sure to do this computation in the right way, mainly
because I'm not sure to understand what is computed in the torque vector
returned by ODE. Basically, I have theses questions:
- do I have to add the two contributions ?
- do I use the good angular velocity ?
- I get negative values for _power, does it mean that the system get
some new energy?
And, of course, is it the "good way" to compute the energy ?
I understand that these questions are more physics-related than
ODE-related but I'm sure the answers will interest some other ODE users
as well.
Jean-Baptiste Mouret / Mandor
tel : (+33) 6 28 35 10 49
More information about the ODE mailing list | {"url":"http://ode.org/old_list_archives/2006-September/020180.html","timestamp":"2014-04-18T23:16:36Z","content_type":null,"content_length":"3942","record_id":"<urn:uuid:096ec54d-029b-46cb-b269-fa73864ac16a>","cc-path":"CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00214-ip-10-147-4-33.ec2.internal.warc.gz"} |
Physics Course Descriptions
Physics 100 - General Physics I
Full course for one year. Fall semester: calculus-based introduction to the classical mechanics of particles and systems—kinematics, laws of motion, conservation principles, rotational dynamics,
oscillators, gravitation. Spring semester: electricity and magnetism, optics, and other topics at the discretion of the instructor. Corequisite: Mathematics 111 or equivalent.
Physics 200 - General Physics II
Full course for one year. Fall semester: AC circuits, damped and driven vibrations, coupled oscillators, waves. Related mathematical methods are introduced: complex numbers, ordinary differential
equations, linear algebra, and Fourier analysis. Weekly laboratories provide an introduction to basic electronics, from filters and voltage dividers to transistors and operational amplifiers as well
as damped, driven, and coupled oscillators. Spring semester: thermal physics, modern physics—introduction to special relativity and quantum mechanics, with applications to atomic, nuclear, and
particle physics, and condensed matter, as time permits. Weekly laboratories include an introduction to Mathematica, the Millikan oil drop experiment, measurement of the speed of light, determination
of Planck’s constant, the charge-to-mass ratio of the electron, microwaves and high-temperature superconductors. Prerequisites: Physics 100; Mathematics 111 (or equivalent) and 112; Mathematics
211-212 should be taken concurrently. First-year students who have successfully completed the equivalent of Physics 100 at the college level may petition the physics department to take Physics 200 in
their first year. The petition must offer evidence of proficiency in calculus-based electricity and magnetism. Lecture-conference-laboratory.
Physics 311 - Classical Mechanics I
Full course for one semester. Careful examination of the foundations and limitations of Newtonian mechanics leads to development of the Lagrangian formulation, variational principles, Hamiltonian
mechanics, and the theory of canonical transformations. Applications to the motion of rigid bodies, systems of coupled oscillators, and celestial mechanics are treated as time permits. Prerequisite:
Physics 200. Lecture.
Physics 321 - Electrodynamics I
Full course for one semester. Electrostatics and magnetostatics in vacuum and in matter, electromagnetic induction, force and energy in electrodynamics, Maxwell’s equations. Mathematical methods
introduced include multivariable calculus and the solution of partial differential equations by separation of variables. Prerequisite: Physics 200. Lecture.
Physics 322 - Electrodynamics II
Full course for one semester. A continuation of Physics 321, this course emphasizes time-varying electric and magnetic fields. Topics include radiation from point charges and dipoles; propagation of
electromagnetic plane waves in vacuum and in matter; reflection, refraction, and dispersion; and the relativistic formulation of electrodynamics. Prerequisite: Physics 321. Lecture.
Physics 323 - Optics
Full course for one semester. Theories of light, from the 17th century to the present. Electromagnetic theory and the modern photon picture. Applications of geometrical optics, including lenses,
prisms, polarizers, wave plates; reflection and refraction in general. Huygens’ Principle, Fermat’s Principle, diffraction and holography, introduction to quantum optics. Prerequisite: Physics 200.
Physics 331 - Advanced Laboratory I
One-half course for one semester. A study of advanced electronics and computer-assisted data acquisition and analysis intended to provide the student with a basis for understanding and designing
laboratory systems used in contemporary experimental physics. Topics include operational amplifiers, filters, oscillators, logic circuits, and computer interfacing and analysis using a LabVIEW
system. Prerequisite: Physics 200. Lecture-laboratory.
Physics 332 - Advanced Laboratory II
One-half course for one semester. Guided and independent experimental investigations of physical phenomena using research-style measurement techniques. Prerequisite: Physics 331. Lecture-laboratory.
Physics 342 - Quantum Mechanics I
Full course for one semester. An introduction to quantum theory, beginning with the Schrödinger equation and the statistical interpretation of the wave function. One-dimensional applications,
including the infinite square-well, the harmonic oscillator, and scattering; in three dimensions, the theory of angular momentum, central potentials, and the hydrogen atom; time-independent
perturbation theory, spin, identical particles, and the Pauli exclusion principle. In general, this course concentrates on exact solutions to artificial problems, in contrast to Quantum Mechanics II,
which develops approximate solutions to real problems. Prerequisite: Physics 200. Lecture.
Physics 351 - Thermal Physics
Full course for one semester. Examines the essentials of probability and statistics, the kinetic theory of gases, statistical mechanics, temperature, equations of state, heat, internal energy,
entropy, reversibility, and distribution functions. Prerequisite: Physics 200. Lecture.
Physics 362 - Solid State Physics
Full course for one semester. Crystalline lattice structures, vibrational modes, and electronic band theory are explored and used to explain the observed electrical, thermal, optical, and magnetic
properties of solids. Prerequisite: Physics 200. Lecture. Not offered 2008-09.
Physics 363 - Molecular Biophysics
Full course for one semester. The course will cover the physics ofmeasurement techniques for studying the most significant intermolecularinteractions of synaptic transmission. An introduction to the
biologyof neurons will be provided. Measurement techniques such as evanescentwave microscopy, confocal microscopy, X-ray diffraction, fluorescenceresonance energy transfer, and Raman and infrared
spectroscopy will beexplained in terms of the physics of the experiment and itsimplementation. A clear idea of how these measurements inform themodels of cellular processes such as exocytosis as well
as the atomic-level models of neuromolecular structure and function will bepresented. The course will include demonstrations of selectedmeasurement techniques such as total internal reflection
microscopy,infrared absorption, and crystallography. Prerequisites: Physics 100,Physics 200, Mathematics 211 and 212. Lecture.
Physics 364 - Selected Topics of Astrophysical Interest
Full course for one semester. Specific topics vary from year to year, drawn principally from the following areas: internal constitution, evolution, and death of stars; structure of galaxies;
interstellar medium; radiative processes; and classical cosmology. Prerequisite: Physics 200. Lecture.
Physics 366 - Elementary Particles
Full course for one semester. Introduction to the theory and phenomenology of elementary particle physics. The course includes a semihistorical overview, followed by relativistic kinematics, the
Dirac equation, evaluation of simple Feynman diagrams, and a survey of the strong, electromagnetic, and weak interactions from the perspective of gauge theory. Prerequisite: Physics 200.
Lecture-conference. Not offered 2008-09.
Physics 367 - Scientific Computation
Full course for one semester. This course covers numerical and laboratory methods for students of science. The primary focus will be on topics in physics, chemistry, and biology. The course begins
with the history and modern importance of scientific computation, moves on to methodology and specific algorithms, and closes with individual elective projects to be approved by the instructor. Basic
programming will not be taught; the course will concentrate on scientific, not programmatic, aspects, so students must be able to write programs largely on their own. Specific topics include
differential equations, matrix methods, signal and image processing, quantum-theoretic models, astrophysical models, and nonlinear and chaotic systems. Prerequisites: a sophomore-level course in one
of the sciences and experience with a sufficiently strong computer language, such as Pascal or C. Lecture-conference-laboratory. Cross-listed as Biology 367.
Physics 411 - Classical Mechanics II
Full course for one semester. A continuation of Physics 311; specific content varies from year to year. Prerequisite: Physics 311. Lecture-conference.
Physics 414 - Classical Field Theory
Full course for one semester. A modern account of the classical dynamics of systems with infinitely many degrees of freedom. Treats both general principles and more specialized techniques appropriate
to the analysis of topics of exceptional current interest (solitons, gauge fields). Although primarily for physicists, the course contains much material of interest to mathematicians. A good command
of classical mechanics, linear algebra, and the theory of differential equations is assumed. Lecture. Not offered 2008-09.
Physics 442 - Quantum Mechanics II
Full course for one semester. A continuation of Physics 342; specific content varies from year to year. The emphasis is on approximation techniques (time-independent and time-dependent perturbation
theory, WKB approximation, variational principles, Born approximation), with applications to atoms, molecules, and solids, the quantum theory of radiation, and formal scattering theory. Prerequisite:
Physics 342. Lecture.
Physics 470 - Thesis and Physics Seminar
Full course for one year. The thesis is independent work on an original problem and is intended as an introduction to research. In addition to the thesis project itself, all seniors are expected to
participate in a weekly seminar in which various topics from the current literature are discussed.
Physics - Special Topics in Physics
One-half or full course for one semester. Readings and laboratory work of an advanced character. Students will choose a field in which they are interested; they are expected to become familiar with
the special instruments and methods of that discipline. Open only to juniors and seniors, by consent of the instructor. Lecture-conference. | {"url":"http://www.reed.edu/catalog/archives/catalog_2008_09/courses/phys/index.html","timestamp":"2014-04-20T13:21:57Z","content_type":null,"content_length":"23199","record_id":"<urn:uuid:5be7ed4a-2640-4c12-8569-07207b620389>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00039-ip-10-147-4-33.ec2.internal.warc.gz"} |
Physics Department, Princeton University
Events - Daily
Friday, May 10
High Energy Theory Seminar - IAS - Thomas Faulkner, IAS - “The Entanglement Renyi Entropies of Disjoint Intervals in AdS/CFT”
I will discuss how to compute Entanglement Renyi Entropies for two dimensional holographic CFTs using AdS_3 handlebody solutions. Some simplifying assumptions are made. The answer I find is
consistent with the Ryu-Takayanagi prediction for Entanglement Entropy.
Bloomberg Lecture Hall - Institute for Advanced Study · 1:30 p.m.– 2:30 p.m. | {"url":"http://www.princeton.edu/physics/events_archive/calendar_daily.xml?displayday=2013-05-10","timestamp":"2014-04-21T12:42:05Z","content_type":null,"content_length":"10068","record_id":"<urn:uuid:da07ea7a-8044-4e54-a21f-adb908f797cd>","cc-path":"CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00533-ip-10-147-4-33.ec2.internal.warc.gz"} |
Reverse Calculation Of Formula
how can i create a formula by 3 variable?( power form: y=(x^a)*(v^b)*(g^k)*p that y,x,v,g are known and other parameter are unknown.)
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Here is the formula:
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D E F G H I J
Loan Dsc. Int. DSCNT% Y-S-P% Fees Loan Amount Commission
80/20 7.52.0 1.0 3.0 $137,403 $4,122.09
I attempted this formula and obviously it’s incorrect:
Calculating Commission (J2)
=IF(H2=" "," ",=IF(D2="80/20",(I2*0.80)*(H2*0.01),I2 * (H2*0.01))
1. If H2 is blank then TRUE enter a blank
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Cumberland SW
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A while back I posted a question about this sort of reverse two way lookup:
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If it is easier to see the table with dates and times, the data is here:
Schedule looks like this:
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etd AT Indian Institute of Science: Statistical Mechanical Models Of Some Condensed Phase Rate Processes
& Collections
Thesis Guide
Submitted Date
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etd AT Indian Institute of Science >
Division of Chemical Sciences >
Inorganic and Physical Chemistry (ipc) >
Please use this identifier to cite or link to this item: http://hdl.handle.net/2005/863
Title: Statistical Mechanical Models Of Some Condensed Phase Rate Processes
Authors: Chakrabarti, Rajarshi
Advisors: Sebastian, K L
Statistical Mechanics
Transition State Theory
Kramers Theory
Keywords: Reactive Flux
Proteins - Diffusion
Transient State Work Fluctuation Theorem
Rate Processes
First Passage Time Theory
Non-Ohmic Bath
Submitted Sep-2008
Series/ G22592
Report no.:
In the thesis work we investigate four problems connected with dynamical processes in condensed medium, using different techniques of equilibrium and non-equilibrium statistical
mechanics. Biology is rich in dynamical events ranging from processes involving single molecule [1] to collective phenomena [2]. In cell biology, translocation and transport processes of
biological molecules constitute an important class of dynamical phenomena occurring in condensed phase. Examples include protein transport through membrane channels, gene transfer
between bacteria, injection of DNA from virus head to the host cell, protein transport thorough the nuclear pores etc. We present a theoretical description of the problem of protein
transport across the nuclear pore complex [3]. These nuclear pore complexes (NPCs) [4] are very selective filters that monitor the transport between the cytoplasm and the nucleoplasm.
Two models have been suggested for the plug of the NPC. The first suggests that the plug is a reversible hydrogel while the other suggests that it is a polymer brush. In the thesis, we
propose a model for the transport of a protein through the plug, which is treated as elastic continuum, which is general enough to cover both the models. The protein stretches the plug
and creates a local deformation, which together with the protein is referred to as the bubble. The relevant coordinate describing the transport is the center of the bubble. We write down
an expression for the energy of the system, which is used to analyze the motion. It shows that the bubble executes a random walk, within the gel. We find that for faster relaxation of
the gel, the diffusion of the bubble is greater. Further, on adopting the same kind of free energy for the brush too, one finds that though the energy cost for the entry of the particle
is small but the diffusion coefficient is much lower and hence, explanation of the rapid diffusion of the particle across the nuclear pore complex is easier within the gel model. In
chemical physics, processes occurring in condensed phases like liquid or solid often involve barrier crossing. Simplest possible description of rate for such barrier crossing phenomena
is given by the transition state theory [5]. One can go one step further by introducing the effect of the environment by incorporating phenomenological friction as is done in Kramer’s
theory [6]. The “method of reactive flux” [7, 8] in chemical physics allows one to calculate the time dependent rate constant for a process involving large barrier by expressing the rate
as an ensemble average of an infinite number of trajectories starting at the barrier top and ending on the product side at a specified later time. We compute the time dependent
Abstract: transmission coefficient using this method for a structureless particle surmounting a one dimensional inverted parabolic barrier. The work shows an elegant way of combining the
traditional system plus reservoir model [9] and the method of reactive flux [7] and the normal mode analysis approach by Pollak [10] to calculate the time dependent transmission
coefficient [11]. As expected our formula for the time dependent rate constant becomes equal to the transition state rate constant when one takes the zero time limit. Similarly Kramers
rate constant is obtained by taking infinite time limit. Finally we conclude by noting that the method of analyzing the coupled Hamiltonian, introduced by Pollak is very powerful and it
enables us to obtain analytical expressions for the time dependent reaction rate in case of Ohmic dissipation, even in underdamped case. The theory of first passage time [12] is one of
the most important topics of research in chemical physics. As a model problem we consider a particle executing Brownian motion in full phase space with an absorbing boundary condition at
a point in the position space we derive a very general expression of the survival probability and the first passage time distribution, irrespective of the statistical nature of the
dynamics. Also using the prescription adopted elsewhere [13] we define a bound to the actual survival probability and an approximate first passage time distribution which are expressed
in terms of the position-position, velocity-velocity and position-velocity variances. Knowledge of these variances enables one to compute the survival probability and consequently the
first passage distribution function. We compute both the quantities for gaussian Markovian process and also for non-Markovian dynamics. Our analysis shows that the survival probability
decays exponentially at the long time, irrespective of the nature of the dynamics with an exponent equal to the transition state rate constant [14]. Although the field of equilibrium
thermodynamics and equilibrium statistical mechanics are well explored, there existed almost no theory for systems arbitrarily far from equilibrium until the advent of fluctuation
theorems (FTs)[15] in mid 90�s. In general, these fluctuation theorems have provided a general prescription on energy exchanges that take place between a system and its surroundings
under general nonequilibrium conditions and explain how macroscopic irreversibility appears naturally in systems that obey time reversible microscopic dynamics. Based on a Hamiltonian
description we present a rigorous derivation [16] of the transient state work fluctuation theorem and the Jarzynski equality [17] for a classical harmonic oscillator linearly coupled to
a harmonic heat bath, which is dragged by an external agent. Coupling with the bath makes the dynamics dissipative. Since we do not assume anything about the spectral nature of the
harmonic bath the derivation is valid for a general non-Ohmic bath.
URI: http://hdl.handle.net/2005/863
Appears in Inorganic and Physical Chemistry (ipc)
Items in etd@IISc are protected by copyright, with all rights reserved, unless otherwise indicated. | {"url":"http://etd.ncsi.iisc.ernet.in/handle/2005/863","timestamp":"2014-04-18T03:42:13Z","content_type":null,"content_length":"25521","record_id":"<urn:uuid:2d3e538e-24db-4ccb-af0a-5b68409510f7>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00481-ip-10-147-4-33.ec2.internal.warc.gz"} |
Table of Contents | Introduction | Home
P l a n e G e o m e t r y
An Adventure in Language and Logic
based on
Book I. Proposition 8
SUPPOSE THAT two sides of one triangle are equal respectively to
two sides of another. What will be a sufficient condition for the angles that are contained by those sides to be equal, the angles A and D? Because, if those angles are equal, then the triangles will
be congruent, Side-angle-side.
A sufficient condition for those angles to be equal . . . is that the bases are also equal.
Again, Euclid proved this by superposition. And again we will accept it as a postulate.
(As for Proposition 7, it was a theorem Euclid needed to prove Proposition 8 by superposition.)
If two triangles have two sides equal to two sides respectively, and if the bases are also equal, then those angles will be equal that are contained by the two equal sides.
Let triangles ABC, DEF be two triangles, with the two sides AB, AC equal to the two sides DE, DF respectively;
and let the base BC be equal to the base EF;
then angle BAC will equal angle EDF.
Since those angles are equal, then the triangles are congruent
according to S.A.S. Nevertheless, we will say that this gives a second condition for triangles to be congruent, and we will often refer to this proposition as "S.S.S.": Side-Side-Side.
In the enunciation, it would be redundant to state "the triangles themselves will be equal areas, and the remaining angles will be equal, namely those that are opposite the equal sides," because we
have already stated that in S.A.S. See the following Example.
Example. In this figure, AB is equal to DC, and AC is equal to DB. Prove that angle BAC is equal to angle CDB, and that angle ABC is equal to angle DCB.
Solution. Since AB is equal to DC, and AC to DB,
then the two sides BA, AC are equal to the two sides CD, DB respectively;
and the base BC is common to triangles BAC, CDB;
therefore, angle BAC is equal to angle CDB (S.S.S.).
Therefore the triangles are congruent,
and those angles are equal that are opposite the equal sides. (S.A.S.)
Therefore angle ABC, opposite side CA, is equal to angle DCB, opposite the equal side BD.
Please "turn" the page and do some Problems.
Continue on to the next proposition.
Previous proposition
Table of Contents | Introduction | Home
Please make a donation to keep TheMathPage online.
Even $1 will help.
Copyright © 2014 Lawrence Spector
Questions or comments?
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Strategy for the Prediction and Selection of Drug Substance Salt Forms - Pharmaceutical Technology
then Equation 22 becomes:
When converted to the Sørensen scale, Equation 23 becomes:
Equation 25 can be used to make rapid deductions regarding the strength of a particular salt species. Suppose one were contemplating forming a salt between an acid having a pK [ A ] value of 4.79 and
a base having a pK [ A ] value of 9.45. For the base, it would follow that the pK [ B ] would equal 4.55, and because pK [ W ] = 14.0 at 25 °C, the pK [ S ] of the salt would equal -4.66, and that K
[ S ] would equal approximately 45,710 . A reaction characterized by an equilibrium constant of this magnitude would clearly go to completion, and one would predict that the salt in question would be
formed without difficulty.
The ability to calculate K [ S ] enables one to estimate the relative position of the equilibrium described by Equation 16. Consider the solution prepared by mixing an acid at an initial
concentration of C [HA] with a base at an initial concentration of C [ B ]. For a salt form having a 1:1 stoichiometry, the concentrations of conjugate acid and conjugate base formed in the reaction
would necessarily be equal. If the resulting ionic concentrations are represented by X, then the concentration of residual acid would equal (C [HA]–X) and the concentration of residual base would
equal (C [ B ]–X). Equation 24 would then have the form:
Because obtaining the solution of Equation 26 by means of the quadratic equation is trivial, the degree of formation of a salt through the mixing of equimolar amounts of acid and base (i.e., C [HA] =
C [ B ]) can be easily calculated. For example, if log(K [ S ]) = 2, it follows that X = 0.9091, indicating that equation 16 would proceed 90.91% to completion. Similarly, if log(K [ S ]) = 3, then X
= 0.9693 and the efficiency of salt formation would be 96.93%. If log(K [ S ]) = 4, the salt would be 99.01% formed, and if log(K [ S ]) = 5, then the salt would be 99.68% formed. It is often stated
in the literature that if the ionization constants of the acid and base involved in salt formation differ by 2 or 3 pK units, then the salt would be formed. Use of the log(K [ S ]) quantity serves to
place the old empirical rule on a more fundamental basis and facilitates calculation of the actual percentage of salt formation.
Knowledge of the log(K [ S ]) quantity also permits one to deduce the degree of disproportionation that would be anticipated if one were to dissolve a salt in pure water. If the X factor of Equation
26 represents the fraction of salt being formed by the reaction of the acid and base, then it follows that the fraction of salt that would disproportionate would necessarily be given by (1–X), and
its percentage as 100 times that quantity. | {"url":"http://www.pharmtech.com/pharmtech/article/articleDetail.jsp?id=463704&sk=&date=&pageID=5","timestamp":"2014-04-16T07:28:43Z","content_type":null,"content_length":"150115","record_id":"<urn:uuid:9ffbc418-4b4e-49d2-a447-b781fbc06eda>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00051-ip-10-147-4-33.ec2.internal.warc.gz"} |
4-genus of a 2-bridge link
up vote 4 down vote favorite
How can we calculate the 4-genus of a link L? The 4-genus is defined to be the minimal genus of orientable surface bounded by L in B^4. Is there any routine method to calculate that?
Especially, any good idea how to calculate the 4-genus of a 2-bridge link? Thanks.
knot-theory gt.geometric-topology 4-manifolds 2-knots
add comment
4 Answers
active oldest votes
You might get interesting lower bounds using Rasmussen's s-invariant, and his calculations of the KR homology of 2-bridge links.
up vote 4 down Or you might not. As Ryan says, there are various ways of bounding genus above and below, but unless you get lucky and two of them match, it can be hard to tell. The 4-genus of
vote accepted torus knots was an open problem for a scandalously long time (if my history is right, it was first proven in the 90's!).
Thanks. I think KR homology is more interesting because it can give a finer lower bound for 4-genus. As for Rasmussen's s-invariant, it equals the signature when the link is
alternating, which seems kind of trivial. – Megan Feb 9 '10 at 5:05
1 I completely disagree. Remember, there aren't that many alternating knots. Separating knots into alternating and not-alternating is like separating the world into bananas and
not-bananas, to steal a famous quote. – Ben Webster♦ Feb 9 '10 at 5:23
add comment
Take a look at Paolo Lisca's papers "Rational balls and the ribbon conjecture" and "Sums of lens spaces bounding rational balls". He determines which 2-bridge knots are slice and the
concordance order for all 2-bridge knots.
up vote 4
down vote But no there's no known effective procedure to compute 4-genus.
Thanks a lot. They are really helpful but don't apply to the case I'm interested in. It seems that not the concordance order of ALL 2-bridges knots K(p,q) is determined but only for
those with p odd. And the knot I'm studying has p=208, which is neither odd nor a square. – Megan Feb 7 '10 at 21:06
add comment
Rasmussen and Lee's results say that the $s$ invariant of a 2-bridge knot will be just equal to the signature of the knot. So you can compute the signature of your knot to get a lower bound
(there are extremely rapid ways of doing this from an alternating diagram). Unfortunately the only decent way to get an upper bound that I know of is by spotting a smooth surface! Good
up vote 3
down vote Remember that $s$ might not be the best you can do. In particular, among alternating knots, the figure 8 knot ($4_1$ in Rolfsen) has vanishing $s$ invariant (for example, because it is
torsion in the concordance group) and yet it is not even slice if you allow your surfaces to be locally flat, let alone smooth.
Aha! Thanks. I'm reading your paper "a slice genus lower bound from $sl(n)$ KR homology" these days. – Megan Feb 10 '10 at 18:08
That's nice! However, I think that you won't get any better bounds on 2-bridge knots with the KR bounds than with the signature or with standard $s$ or $\tau$. This is because if the
bounds are a homomorphism to Z from the concordance group and are tight for the positive knots, it forces them all to agree for alternating knots. So the possibility that the KR bounds
are better since the $E_2$ homology doesn't have to be thin is not really a possibility after all... – Andrew Lobb Feb 10 '10 at 18:46
Thanks. I'm not quite sure what you denote by $\tau$. Is it the Thurston-Bennequin invariant? I thought KR might be a better one since it's more sophisticated than many other
invariants.But it may be a good news for me that KR doesn't give a better bound. I don't need to be involved in the tough computation of KR any more. – Megan Feb 11 '10 at 6:08
Sorry - I meant the bound on the slice genus coming from Heegaard-Floer knot homology. – Andrew Lobb Feb 13 '10 at 0:59
add comment
(This is really a comment on the answer relating to concordance order.)
Since your p is even, then your 2-bridge knot is actually a link. So, while it makes sense to ask if it's a slice or ribbon link, asking about its concordance order doesn't make
up vote 2 down vote sense.
add comment
Not the answer you're looking for? Browse other questions tagged knot-theory gt.geometric-topology 4-manifolds 2-knots or ask your own question. | {"url":"http://mathoverflow.net/questions/14530/4-genus-of-a-2-bridge-link?answertab=votes","timestamp":"2014-04-19T10:35:49Z","content_type":null,"content_length":"71516","record_id":"<urn:uuid:67793851-ed0e-4c63-aa57-160061c68abe>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00372-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Basic Differentiation Rules
Some differentiation rules are a snap to remember and use. These include the constant rule, power rule, constant multiple rule, sum rule, and difference rule.
• The constant rule: This is simple. f (x) = 5 is a horizontal line with a slope of zero, and thus its derivative is also zero.
• The power rule:
To repeat, bring the power in front, then reduce the power by 1. That’s all there is to it.
The power rule works for any power: a positive, a negative, or a fraction.
Make sure you remember how to do the last function. It’s the simplest function, yet the easiest problem to miss. By the way, do you see how finding this last derivative follows the power rule?
(Hint: x to the zero power equals one).
You can differentiate radical functions by rewriting them as power functions and then using the power rule.
• The constant multiple rule: What if the function you’re differentiating begins with a coefficient? Makes no difference. A coefficient has no effect on the process of differentiation. You just
ignore it and differentiate according to the appropriate rule. The coefficient stays where it is until the final step when you simplify your answer by multiplying by the coefficient.
Don’t forget that ð (~3.14) and e (~2.72) are numbers, not variables, so they behave like ordinary numbers. Constants in problems, like c and k, also behave like ordinary numbers. So, for
• The sum rule: When you want the derivative of a sum of terms, take the derivative of each term separately.
• The difference rule: If you have a difference (that’s subtraction) instead of a sum, it makes no difference. You still differentiate each term separately.
The addition and subtraction signs are unaffected by the differentiation. | {"url":"http://www.dummies.com/how-to/content/the-basic-differentiation-rules.navId-403861.html","timestamp":"2014-04-18T19:06:37Z","content_type":null,"content_length":"54867","record_id":"<urn:uuid:29aba55b-ea48-4d58-81f8-7678e47499b1>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00077-ip-10-147-4-33.ec2.internal.warc.gz"} |
Rutherford, NJ SAT Math Tutor
Find a Rutherford, NJ SAT Math Tutor
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On Moschovakis Closure Ordinals
Results 1 - 10 of 52
, 1999
"... this paper, we extend XML DTDs with several classes of integrity constraints and investigate the complexity of reasoning about these constraints. The constraints range over keys, foreign keys,
inverse constraints as well as ID constraints for capturing the semantics of object identities. They imp ..."
Cited by 83 (12 self)
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this paper, we extend XML DTDs with several classes of integrity constraints and investigate the complexity of reasoning about these constraints. The constraints range over keys, foreign keys,
inverse constraints as well as ID constraints for capturing the semantics of object identities. They improve semantic specifications and provide a better reference mechanism for native XML
applications. They are also useful in information exchange and data integration for preserving the semantics of data originating in relational and object-oriented databases. We establish complexity
and axiomatization results for the (finite) implication problems associated with these constraints. In addition, we study implication of more general constraints, such as functional, inclusion and
inverse constraints defined in terms of navigation paths
- Information and Computation , 1989
"... A theory satisfies the k-variable property if every first-order formula is equivalent to a formula with at most k bound variables (possibly reused). Gabbay has shown that a model of temporal
logic satisfies the k-variable property for some k if and only if there exists a finite basis for the tempora ..."
Cited by 76 (5 self)
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A theory satisfies the k-variable property if every first-order formula is equivalent to a formula with at most k bound variables (possibly reused). Gabbay has shown that a model of temporal logic
satisfies the k-variable property for some k if and only if there exists a finite basis for the temporal connectives over that model. We give a model-theoretic method for establishing the k-variable
property, involving a restricted Ehrenfeucht-Fraisse game in which each player has only k pebbles. We use the method to unify and simplify results in the literature for linear orders. We also
establish new k-variable properties for various theories of bounded-degree trees, and in each case obtain tight upper and lower bounds on k. This gives the first finite basis theorems for
branching-time models of temporal logic. 1 Introduction A first-order theory \Sigma satisfies the k-variable property if every first-order formula is equivalent under \Sigma to a formula with at most
k bound variables (pos...
- Information and Computation , 1998
"... ..."
, 1998
"... We present a class of path constraints of interest in connection with both structured and semistructured databases, and investigate their associated implication problems. These path constraints
are capable of expressing natural integrity constraints that are not only a fundamental part of the semant ..."
Cited by 63 (18 self)
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We present a class of path constraints of interest in connection with both structured and semistructured databases, and investigate their associated implication problems. These path constraints are
capable of expressing natural integrity constraints that are not only a fundamental part of the semantics of the data, but are also important in query optimization. We show that in semistructured
databases, despite the simple syntax of the constraints, their associated implication problem is r.e. complete and finite implication problem is co-r.e. complete. However, we establish the
decidability of the implication problems for several fragments of the path constraint language, and demonstrate that these fragments suffice to express important semantic information such as inverse
relationships and local database constraints commonly found in object-oriented databases. We also show that in the presence of types, the analysis of path constraint implication becomes more
delicate. We demonst...
, 1990
"... In this paper we ask the question, "What must be added to first-order logic plus least-fixed point to obtain exactly the polynomial-time properties of unordered graphs?" We consider the
languages Lk consisting of first-order logic restricted to k variables and Ck consisting of Lk plus "counting ..."
Cited by 57 (7 self)
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In this paper we ask the question, "What must be added to first-order logic plus least-fixed point to obtain exactly the polynomial-time properties of unordered graphs?" We consider the languages Lk
consisting of first-order logic restricted to k variables and Ck consisting of Lk plus "counting quantifiers". We give efficient canonization algorithms for graphs characterized by Ck or Lk . It
follows from known results that all trees and almost all graphs are characterized by C2 .
- Information and Computation , 1995
"... The extensions of first-order logic with a least fixed point operator (FO + LFP) and with a partial fixed point operator (FO + PFP) are known to capture the complexity classes P and PSPACE
respectively in the presence of an ordering relation over finite structures. Recently, Abiteboul and Vianu [Abi ..."
Cited by 56 (6 self)
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The extensions of first-order logic with a least fixed point operator (FO + LFP) and with a partial fixed point operator (FO + PFP) are known to capture the complexity classes P and PSPACE
respectively in the presence of an ordering relation over finite structures. Recently, Abiteboul and Vianu [Abiteboul and Vianu, 1991b] investigated the relationship of these two logics in the
absence of an ordering, using a machine model of generic computation. In particular, they showed that the two languages have equivalent expressive power if and only if P = PSPACE. These languages can
also be seen as fragments of an infinitary logic where each formula has a bounded number of variables, L ! 1! (see, for instance, [Kolaitis and Vardi, 1990]). We investigate this logic of finite
structures and provide a normal form for it. We also present a treatment of the results in [Abiteboul and Vianu, 1991b] from this point of view. In particular, we show that we can write a formula of
FO + LFP that defines ...
- JOURNAL OF COMPUTER AND SYSTEM SCIENCES , 1990
"... We study here the language Datalog(6=), which is the query language obtained from Datalog by allowing equalities and inequalities in the bodies of the rules. We view Datalog(6=) as a fragment of
an innitary logic L ! and show that L ! can be characterized in terms of certain two-person pebble ga ..."
Cited by 52 (9 self)
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We study here the language Datalog(6=), which is the query language obtained from Datalog by allowing equalities and inequalities in the bodies of the rules. We view Datalog(6=) as a fragment of an
innitary logic L ! and show that L ! can be characterized in terms of certain two-person pebble games. This characterization provides us with tools for investigating the expressive power of Datalog(6
=). As a case study, we classify the expressibility of fixed subgraph homeomorphism queries on directed graphs. Fortune et al. [FHW80] classied the computational complexity of these queries by
establishing two dichotomies, which are proper only if P 6= NP. Without using any complexity-theoretic assumptions, we show here that the two dichotomies are indeed proper in terms of expressibility
in Datalog(6=).
- Journal of Computer and System Sciences , 1996
"... We study topological queries over two-dimensional spatial databases. First, we show that the topological properties of semi-algebraic spatial regions can be completely specified using a
classical finite structure, essentially the embedded planar graph of the region boundaries. This provides an invar ..."
Cited by 45 (2 self)
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We study topological queries over two-dimensional spatial databases. First, we show that the topological properties of semi-algebraic spatial regions can be completely specified using a classical
finite structure, essentially the embedded planar graph of the region boundaries. This provides an invariant characterizing semi-algebraic regions up to homeomorphism. All topological queries on
semi-algebraic regions can be answered by queries on the invariant whose complexity is polynomially related to the original. Also, we show that for the purpose of answering topological queries,
semi-algebraic regions can always be represented simply as polygonal regions. We then study query languages for topological properties of two-dimensional spatial databases, starting from the
topological relationships between pairs of planar regions introduced by Egenhofer. We show that the closure of these relationships under appropriate logical operators yields languages which are
complete for topological prope...
- Information and Computation , 1992
"... We investigate the in nitary logic L 1! , in which sentences may have arbitrary disjunctions and conjunctions, but they involve only a nite number of distinct variables. We show that various
xpoint logics can be viewed as fragments of L 1! , and we describe a game-theoretic characterizat ..."
Cited by 43 (4 self)
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We investigate the in nitary logic L 1! , in which sentences may have arbitrary disjunctions and conjunctions, but they involve only a nite number of distinct variables. We show that various xpoint
logics can be viewed as fragments of L 1! , and we describe a game-theoretic characterization of the expressive power of the logic. Finally, we study asymptotic probabilities of properties 1! on nite
structures. We show that the 0-1 law holds for L 1! , i.e., the asymptotic probability of every sentence in this logic exists and is equal to either 0 or 1. This result subsumes earlier work on
asymptotic probabilities for various xpoint logics and reveals the boundary of 0-1 laws for in nitary logics. | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=1230169","timestamp":"2014-04-16T13:11:20Z","content_type":null,"content_length":"36138","record_id":"<urn:uuid:8aa14cf4-de1c-449c-abf5-699f762e585b>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00199-ip-10-147-4-33.ec2.internal.warc.gz"} |
aterfall plot
Waterfall plot
From setiquest wiki
A waterfall plot is a three dimensional graphical display (frequency vs. time vs. intensity) also known as a spectrogram.
Pseudo-code for generating waterfall plots
This pseudo-code omits potential refinements (e.g., windowing to improve channel response).
1. Perform a FFT of size n on the first n complex samples of data.
2. Square the FFT output channels (real^2 + imag^2) to compute power for each of the n channels.
3. Autoscale the n power values into n pixel values.
4. Arrange the n pixel values into the top row of a waterfall image of width n (by convention: most negative frequency left-most, most positive frequency right-most).
5. Perform a FFT of size n on the next n complex samples.
6. Same as step 2.
7. Same as step 3.
8. Same as step 4, placing n pixels in next waterfall image row down (by convention: time flows from top-to-bottom).
9. While there is more data, go to step 5.
The resulting waterfall plot will be n pixels wide and its height will be determined by the number of FFTs that the observation data file length supports (i.e., file length / (2 bytes-per-sample) /
FFT size).
What value should you choose for n, the FFT size? The larger the FFT, the narrower the bandwidth of each FFT output channel which supports greater signal-to-noise ratio because the noise power (which
is conserved) is distributed randomly (evenly) in all the FFT output channels but the power of an embedded narrow-band signal will be wholly contained in just one channel, or a few channels. However,
a large FFT size will reduce transient response and can hide, in combination with auto-scaling, wide features, such as the hydrogen spectrum.
The waterfall plots computed for some of the setiQuest observations have a channel width, or equivalent vertical pixel width, of ~ 1Hz. This is a sweet spot between maximizing signal-to-noise for
narrow band signals and minimizing the effect of frequency spreading by the interstellar medium. The sample rate of the setiQuest data is 8738133.(3) samples/second. Therefore, we use a binary-length
FFT size (n) of 2^23 (8388608). The setiQuest analysis FFT is done in two stages and the output channels are divided into 256 pixel-wide waterfall images.
Auto-scaling of the power values into pixel values is done to achieve nice contrast for signals against background noise. There is no one best technique for doing this. However, a technique which
works well -- the technique used to auto-scale the setiQuest waterfall images -- is to generate statistics on the power values which will approximate a normal distribution for almost all observation
data, even data containing strong signals. Set all power values below one standard deviations to black and all power values above 3 standard deviations to white with all other values mapped linearly
between the two.
See also
• E-mail from Rob Ackermann, 2010-11-30. | {"url":"http://setiquest.org/wiki/index.php?title=Waterfall_plot&oldid=2729","timestamp":"2014-04-19T09:55:31Z","content_type":null,"content_length":"16037","record_id":"<urn:uuid:f36e766d-ac10-42ed-9de5-5197855ee5e0>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00543-ip-10-147-4-33.ec2.internal.warc.gz"} |
Where is the center of the universe?
I feel like you cannot compare the expansion of the universe to ants on a balloon. Ants can only move across the surface of the balloon.
Yes, and we can only move in our three dimensions of space.
Anyway, the main point here is that a center is a point of symmetry. It's a unique point in the space where stuff is the same in basically every direction from it. The Sun, for example, is a
reasonable center for our solar system because if we choose it as our center, the rest of the solar system moves around it. If you're near the Earth, then the center of the Earth makes for a good
center for most things. If you're talking about our galaxy, well, that's got a center too: Sagittarius A*.
What makes all of these important is they are points of symmetry: the galaxy, or the solar system, or the Earth all look very similar if you rotate those respective systems around those centers. The
universe as a whole has no such unique point of symmetry. Instead, you can rotate the entire universe all you like around
point and it will, more or less, look the same. There simply isn't any unique point of symmetry for the whole universe. | {"url":"http://www.physicsforums.com/showthread.php?p=3717905","timestamp":"2014-04-19T22:45:07Z","content_type":null,"content_length":"77073","record_id":"<urn:uuid:79d22c5d-b784-4373-91ba-68e1c8eb1c01>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00284-ip-10-147-4-33.ec2.internal.warc.gz"} |
Bifurcation analysis in a vortex flow generated by an oscillatory magnetic obstacle
Publication: Research - peer-review › Journal article – Annual report year: 2010
Bifurcation analysis in a vortex flow generated by an oscillatory magnetic obstacle. / Beltrán, A.; Ramos, E.; Cuevas, S.; Brøns, Morten.
Physical Review E (Statistical, Nonlinear, and Soft Matter Physics)
, Vol. 81, No. 3, 2010, p. 036309.
Publication: Research - peer-review › Journal article – Annual report year: 2010
Beltrán, A, Ramos, E, Cuevas, S
& Brøns, M
2010, '
Bifurcation analysis in a vortex flow generated by an oscillatory magnetic obstacle
Physical Review E (Statistical, Nonlinear, and Soft Matter Physics)
, vol 81, no. 3, pp. 036309.,
Beltrán, A., Ramos, E., Cuevas, S.
, & Brøns, M.
Bifurcation analysis in a vortex flow generated by an oscillatory magnetic obstacle
Physical Review E (Statistical, Nonlinear, and Soft Matter Physics)
(3), 036309.
title = "Bifurcation analysis in a vortex flow generated by an oscillatory magnetic obstacle",
publisher = "American Physical Society",
author = "A. Beltrán and E. Ramos and S. Cuevas and Morten Brøns",
note = "Copyright 2010 American Physical Society",
year = "2010",
doi = "10.1103/PhysRevE.81.036309",
volume = "81",
number = "3",
pages = "036309",
journal = "Physical Review E (Statistical, Nonlinear, and Soft Matter Physics)",
issn = "1539-3755",
TY - JOUR
T1 - Bifurcation analysis in a vortex flow generated by an oscillatory magnetic obstacle
A1 - Beltrán,A.
A1 - Ramos,E.
A1 - Cuevas,S.
A1 - Brøns,Morten
AU - Beltrán,A.
AU - Ramos,E.
AU - Cuevas,S.
AU - Brøns,Morten
PB - American Physical Society
PY - 2010
Y1 - 2010
N2 - A numerical simulation and a theoretical model of the two-dimensional flow produced by the harmonic oscillation of a localized magnetic field (magnetic obstacle) in a quiescent viscous,
electrically conducting fluid are presented. Nonuniform Lorentz forces produced by induced currents interacting with the oscillating magnetic field create periodic laminar flow patterns that can be
characterized by three parameters: the oscillation Reynolds number, Re-omega, the Hartmann number, Ha, and the dimensionless amplitude of the magnetic obstacle oscillation, D. The analysis is
restricted to oscillations of small amplitude and Ha=100. The resulting flow patterns are described and interpreted in terms of position and evolution of the critical points of the instantaneous
streamlines. It is found that in most of the cycle, the flow is dominated by a pair of counter rotating vortices that switch their direction of rotation twice per cycle. The transformation of the
flow field present in the first part of the cycle into the pattern displayed in the second half occurs via the generation of hyperbolic and elliptic critical points. The numerical solution of the
flow indicates that for low frequencies (v.e. Re-omega = 1), two elliptic and two hyperbolic points are generated, while for high frequencies (v.e. Re-omega = 100), a more complex topology involving
four elliptic and two hyperbolic points appear. The bifurcation map for critical points of the instantaneous streamline is obtained numerically and a theoretical model based on a local analysis that
predicts most of the qualitative properties calculated numerically is proposed.
AB - A numerical simulation and a theoretical model of the two-dimensional flow produced by the harmonic oscillation of a localized magnetic field (magnetic obstacle) in a quiescent viscous,
electrically conducting fluid are presented. Nonuniform Lorentz forces produced by induced currents interacting with the oscillating magnetic field create periodic laminar flow patterns that can be
characterized by three parameters: the oscillation Reynolds number, Re-omega, the Hartmann number, Ha, and the dimensionless amplitude of the magnetic obstacle oscillation, D. The analysis is
restricted to oscillations of small amplitude and Ha=100. The resulting flow patterns are described and interpreted in terms of position and evolution of the critical points of the instantaneous
streamlines. It is found that in most of the cycle, the flow is dominated by a pair of counter rotating vortices that switch their direction of rotation twice per cycle. The transformation of the
flow field present in the first part of the cycle into the pattern displayed in the second half occurs via the generation of hyperbolic and elliptic critical points. The numerical solution of the
flow indicates that for low frequencies (v.e. Re-omega = 1), two elliptic and two hyperbolic points are generated, while for high frequencies (v.e. Re-omega = 100), a more complex topology involving
four elliptic and two hyperbolic points appear. The bifurcation map for critical points of the instantaneous streamline is obtained numerically and a theoretical model based on a local analysis that
predicts most of the qualitative properties calculated numerically is proposed.
UR - http://link.aps.org/doi/10.1103/PhysRevE.81.036309
U2 - 10.1103/PhysRevE.81.036309
DO - 10.1103/PhysRevE.81.036309
JO - Physical Review E (Statistical, Nonlinear, and Soft Matter Physics)
JF - Physical Review E (Statistical, Nonlinear, and Soft Matter Physics)
SN - 1539-3755
IS - 3
VL - 81
SP - 036309
ER - | {"url":"http://orbit.dtu.dk/en/publications/bifurcation-analysis-in-a-vortex-flow-generated-by-an-oscillatory-magnetic-obstacle(36e24390-a82d-4e23-8d27-30f8f43c6e82)/export.html","timestamp":"2014-04-17T13:28:24Z","content_type":null,"content_length":"22829","record_id":"<urn:uuid:4d6ea150-e06a-4602-88d7-5c2452b79dce>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00265-ip-10-147-4-33.ec2.internal.warc.gz"} |
Getting the fractional part of a float without using modf()
up vote 7 down vote favorite
I'm developing for a platform without a math library, so I need to build my own tools. My current way of getting the fraction is to convert the float to fixed point (multiply with (float)0xFFFF, cast
to int), get only the lower part (mask with 0xFFFF) and convert it back to a float again.
However, the imprecision is killing me. I'm using my Frac() and InvFrac() functions to draw an anti-aliased line. Using modf I get a perfectly smooth line. With my own method pixels start jumping
around due to precision loss.
This is my code:
const float fp_amount = (float)(0xFFFF);
const float fp_amount_inv = 1.f / fp_amount;
inline float Frac(float a_X)
return ((int)(a_X * fp_amount) & 0xFFFF) * fp_amount_inv;
inline float Frac(float a_X)
return (0xFFFF - (int)(a_X * fp_amount) & 0xFFFF) * fp_amount_inv;
Thanks in advance!
c++ c math bit-manipulation
Shouldn't fp_amount be 0x10000 instead of 0xFFFF? – Mark Ransom Apr 7 '10 at 20:54
Holy crap. Make it an answer so I can accept it! You just fixed my entire accuracy problem! – knight666 Apr 7 '10 at 21:27
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8 Answers
active oldest votes
If I understand your question correctly, you just want the part after the decimal right? You don't need it actually in a fraction (integer numerator and denominator)?
For float f;:
up vote 13 down vote
accepted f = f-(long)f;
1 An if... then... else... in a math function as often used as this? My cache, it weeps! – knight666 Apr 7 '10 at 18:46
1 This is wrong when f is negative. (You're adding two negative numbers.) You don't need the if at all: f = f - (int) f. If f is negative it'll be subtracting a negative int
that's rounded toward zero. – jamesdlin Apr 7 '10 at 18:52
1 Also, you're assuming that the integer portion of the float fits in an int. – jamesdlin Apr 7 '10 at 19:06
@jamesdlin Thanks, psyched myself out with the negative part. And right you are about the float/integer thing. – Daniel Bingham Apr 7 '10 at 19:10
Oh well. This answer is both the easiest to implement and the fastest (about 20% faster than my method). Accepted! – knight666 Apr 9 '10 at 5:09
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As I suspected, modf does not use any arithmetic per se -- it's all shifts and masks, take a look here. Can't you use the same ideas on your platform?
up vote 5 down vote
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I would recommend taking a look at how modf is implemented on the systems you use today. Check out uClibc's version.
up vote 3 (For legal reasons, it appears to be BSD licensed, but you'd obviously want to double check)
down vote
Some of the macros are defined here.
Why all the bit shifting, is really that much of a speed gain? Or does my little int conversion trick have some issue I'm missing? – Daniel Bingham Apr 7 '10 at 18:41
@Daniel Bingham: likely the latter. Floats may not be encoded in the way you think on the platform you're using, so your masks may be off. @sharth: your link relies on some macros, and
I'm having trouble finding them. Can you try your luck and find definitions for EXTRACT_WORDS, INSERT_WORDS, and GET_HIGH_WORD? – Randolpho Apr 7 '10 at 18:44
Int to float is fine. Float to int is horribly slow. – knight666 Apr 7 '10 at 18:45
@Randolpho: git.uclibc.org/uClibc/tree/libm/math_private.h – sharth Apr 7 '10 at 20:37
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I'm not completly sure, but I think that what you are doing is wrong, since you are only considering the mantissa and forgetting the exponent completely.
You need to use the exponent to shift the value in the mantissa to find the actual integer part.
up vote 2 down vote
For a description of the storage mechanism of 32bit floats, take a look here.
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There's a bug in your constants. You're basically trying to do a left shift of the number by 16 bits, mask off everything but the lower bits, then right shift by 16 bits again. Shifting
up vote 2 is the same as multiplying by a power of 2, but you're not using a power of 2 - you're using 0xFFFF, which is off by 1. Replacing this with 0x10000 will make the formula work as intended.
down vote
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Your method is assuming that there are 16 bits in the fractional part (and as Mark Ransom notes, that means you should shift by 16 bits, i.e. multiply by by 0x1000). That might not be
true. The exponent is what determines how many bit there are in the fractional part.
up vote 0 down To put this in a formula, your method works by calculating (x modf 1.0) as ((x << 16) mod 1<<16) >> 16, and it's that hardcoded 16 which should depend on the exponent - the exact
vote replacement depends on your float format.
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Why go to floating point at all for your line drawing? You could just stick to your fixed point version and use an integer/fixed point based line drawing routine instead - Bresenham's
comes to mind. While this version isn't aliased, I know there are others that are.
up vote 0 down
vote Bresenham's line drawing
You mean anti-aliased? – Victor Engel Jan 7 '13 at 14:08
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Seems like maybe you want this.
float f = something;
up vote 0 down vote float fractionalPart = f - floor(f);
This does not work for negative numbers. – Ton van den Heuvel Jan 7 '13 at 13:30
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Not the answer you're looking for? Browse other questions tagged c++ c math bit-manipulation or ask your own question. | {"url":"http://stackoverflow.com/questions/2594913/getting-the-fractional-part-of-a-float-without-using-modf/2594953","timestamp":"2014-04-21T13:51:36Z","content_type":null,"content_length":"100137","record_id":"<urn:uuid:9bbdce0c-588e-400e-8bcc-0457bfe8db9a>","cc-path":"CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00409-ip-10-147-4-33.ec2.internal.warc.gz"} |
Homework Help
Posted by mysterychicken on Wednesday, March 31, 2010 at 11:36am.
28. Use the Distributive Property to simplify x(4x^2 + x + 4)
Is it 4x^3 + x^2 + 4x?
42. To which set of numbers does 0 not belong?
46. Pi belongs to which group of real numbers?
a. Odd numbers
b. rational numbers
c. irrational numbers
48. Can a number be both irrational and an integer?
a. yes
b. no
c. sometimes
Help! Thanks
• Algebra - bobpursley, Wednesday, March 31, 2010 at 12:29pm
48. No, think about that.
46 right
42 books have been written on what zero means. In poorly written high school texts, the issues are ignored, and they just give you some things to memorize. Commonly, zero is not thought of as a
whole number, nor a positive integer, nor a negative integer, however, it is commonly included in the set of "non-negative integers".
Zero did not exist at all until Descartes started to use it as a place holder in the 1700's, well after math had been invented. He did this as a part of the French Academy's efforts to
standardize math notation.
Before Descartes efforts, 1500 would have been written as 15--, and 30+150 would have been tabulated this way
total 18-
If your math teacher is stupid enough to ask you this question, look it up in your textbook and memorize the textbook's definition for the test.
28. correct
• Algebra - mysterychicken, Wednesday, March 31, 2010 at 1:30pm
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