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inference with functional dependencies
Avi Pfeffer avi@eecs.harvard.edu
Mon, 13 Aug 2001 18:08:08 -0400 (EDT)
Inferring equality between types when there are functional dependencies
seems to be less powerful than I expected. Here's a simple example:
class Eq b => C a b | a -> b
data T a = forall b . C a b => T b
data U a = forall b . C a b => U b
compare :: T a -> U a -> Bool
compare (T x) (U y) = x == y
I expected the compiler (GHC) would be able to deduce that the b type in
the representation of T a and U a must be the same, since both stand in a
C a b relationship, and a functionally determines b in that
relationship. Instead I get the message:
Inferred type is less polymorphic than expected
Quantified type variable `b1' is unified with `b'
When checking a pattern that binds
x :: b
y :: b1
In an equation for function `Test.compare':
Test.compare (T x) (U y) = x == y
Is this expected behavior? Is there a way for me to provide the necessary
hints so that code like this could be accepted?
Avi Pfeffer
Avi Pfeffer avi@eecs.harvard.edu www.eecs.harvard.edu/~avi
Maxwell Dworkin 251
Division of Engineering and Applied Sciences Tel: (617) 496-1876
Harvard University Fax: (617) 496-1066
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Very advanced for me! I'm a complete math noob. I'm willing to pay for help.
March 14th 2013, 04:58 PM #1
Mar 2013
Very advanced for me! I'm a complete math noob. I'm willing to pay for help.
Since I don't know the terminology I am just going to make it into a story problem. :/
I am willing to pay $5 by paypal if I can get it answered in the next hour.
Let's say I have a 10% (This number could vary between 0 and 80) chance when digging a hole that I will find treasure. Each time I dig a hole and do not find treasure I have a +5% higher chance
of finding treasure on the next dig, and this compounds until treasure is found. An example:
10% chance: Hole dug no treasure found.
15% chance: Hole dug no treasure found.
20% chance: Hole dug no treasure found.
25% chance: Hole dug, treasure found!
10% chance: Hole dug... ect. ect.
The chance of finding trasure can not go above 80%, even with the 5% bonus.
What forumla could I use to figure out my overall average chance of finding treasure depending on my starting treasure finding percentage? Obviously when starting at 80% my overall average will
be 80% because the chance can not go above that, but what about for 0-79%?
If anyone figures this out they are so amazing!
Thanks for looking at it even!
Re: Very advanced for me! I'm a complete math noob. I'm willing to pay for help.
Hey eric1982.
This sounds like a Markov model where you can use a Markov chain to model the process. Are you aware of Markov Chains?
Markov chain - Wikipedia, the free encyclopedia
March 14th 2013, 05:24 PM #2
MHF Contributor
Sep 2012
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188 helpers are online right now
75% of questions are answered within 5 minutes.
is replying to Can someone tell me what button the professor is hitting...
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This is the testimonial you wrote.
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Passage/freezing T-cell line calculation problems -
Passage/freezing T-cell line calculation problems - (Jan/10/2010 )
I have a specific questions on how to do the calculations and procedure for passage and freezing of non-adherent immortalized T cells. I have them in 25-ml flasks and I need to do the following:
Passage at concentration of 10e5 cells/ml in 10-ml flasks
Freeze back at concentration of 50x10e5 cells/ml (1ml cells in 1ml media).
I have the cell count for two of them as an example: A = 30 x 10e5 c/ml and B = 5 x 10e5 c/ml
I keep getting confused as to how to get the final concentrations. I know the formula Vol final x concentration final = vol initial x concentration initial. And I can solve for the initial volume I
need to add, but I don't really understand what that volume means. I don't know if I am supposed to spin down the cells and at what volume to resuspend them in. I know this is a basic concept for
technique, but I can't figure out all of the steps accurately, so I would appreciate if someone could explain how to do this in a step-by-step manner.
Thank you for your time!
T-cells are suspension cells right? (I don't use them myself)
Note you may find it easier to use units you are familiar with like 50 x10^5 = 5x10^6 or 5 million.
Also note that in the Ci x Vi= Cf x Vf formula each side is the same as working out "n" in n=c x v. The former formula can be substituted into my examples below, but I have broken it down for ease of
If so - you have a culture of cells growing in a volume of liquid.
Collect the liquid and cells into a tube, take an aliquot for counting (20-100 ul). Remember the volume of cells that you have (e.g. 20 ml)
Count the aliquot. (you gave 30 x10^5 cells per ml in your example; also make sure your numbers are LIVE cells, don't count the dead as they play no part in your culture)
so you have:
n=c x v
n=30 x10^5 cells/ml x 20 ml
n=6x10^7 cells total (60 million)
For passage:
Dilute all the cells to 500,000 cells/ml and aliquot out into new flasks, which will give you a ton of flasks and use a ton of medium too. Using the Ci x Vi etc formula, the Vf number you got is the
number of ml you will have finally. Subtract the initial volume you have (20 ml in my example) and the resulting number is the amount of medium you will need to use.
work out how many cells you need for a flask (e.g. 500,000 cells/ml x 20 ml = 10,000,000 cells) and from that work out the volume required from the original culture:
v=10,000,000 cells/3,000,000 cells/ml
v=3.333 ml
Take 16.667 ml of medium, add it to a new flask and then add 3.333 ml of cell suspension.
For freezing:
Spin the cells down and resuspend at 5 million cells per ml in your freezing mix using the total cell number. This will give lots of frozen tubes (60 million cells/5million cells per ml = 12 tubes),
assuming 1 ml/tube.
Work out how many ml of frozen cells you want (e.g. 4) and work out how many cells you need from the original cell suspension concentration:
n=v x c
n=4 ml x 5 million cells/ml
n=20 million cells.
you have 3 million cells/ml:
v=20 million cells/3 million cells/ml
v= 6.667 ml
Take 6.667 ml of your original culture into a tube, spin down and then add 4 ml of freezing mix.
Thanks, that's really helpful. I just have a follow-up about freezing the cells. If I have a lot of samples (like 14), then it seems difficult to spin down different volumes to get each at the
appropriate concentration for freezing. Since they are all at a different cell count, they wouldn't balance out and therefore couldn't all spin at the same time (it seems trivial, but it takes time
when dealing with many samples). Would there be a calculation if I were to spin down 10ml of each sample (for example) to determine how much media to resuspend in to have the correct final
concentration? I would need to find the amount to resuspend in to have the correct concentration of 5x10e6 cells/ml and add 1ml of resuspended cells to 1ml of freezing media (since media is at 2x
concentration). I hope that made sense.
Thanks again!
You need to count your cells....
Use "number of cells =volume x concentration" (i.e. 10 ml x the number you get from counting in cells/ml)
spin down
Resuspend in appropriate volume to give 5 x 10^6 cells/ml using v= n/c (i.e. volume = number of cells from first calulation/5 x 10^6 cells/ml)
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Binary Tree Array
October 8th, 2011, 05:26 PM #1
Elite Member
Join Date
May 2009
Binary Tree Array
I've suggested a data structure I call a Binary Tree Array:
I think it's novel but if anyone recognises it I would be interested in knowing.
Also I vaguely recognize the 'sweet problem' in the form of a 'Viking problem' or maybe a 'Roman soldier decimation problem' or something. The difference is that instead of passing a sweet a
death sentence is passed and the question is where should you position yourself in the original lineup to be the last man standing walking away unharmed.
So maybe there's a known optimal solution?
Last edited by nuzzle; October 8th, 2011 at 07:39 PM.
Re: Binary Tree Array
Further more to the BTA data structure in my previous post. An example. Say you have an array with 8 elements. Initially the binary tree would look like this,
The numbers reflect the number of array elements below each node of the tree. To remove an array element, say the fifth, you set the corresponding 1 in the leaf row to 0 and adjust all parent
nodes accordingly, like
This makes array element removal an O(log N) operation.
To move M elements forward in the array you walk upwards in the tree from the current element until you find a node where the right child has M elements or more. Then from there you follow the
path downwards that results in skipping M elements. You end up M elements away from the current element. Moving M elements backward is analogical. So moving M elements away from the current
position is an O(log N) operation.
The BTA data structure allows the 'sweet problem' to be solved with O(N * log N) complexity. A solution may look like this,
The root node indicates there's just one element left in the array. Following the path of ones down from the root to the leaf level reveals the fourth child is the winner.
Last edited by nuzzle; October 20th, 2011 at 03:53 AM.
October 11th, 2011, 01:20 AM #2
Elite Member
Join Date
May 2009
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[Numpy-discussion] Style for pad implementation in 'pad' namespace or functions under np.lib
[Numpy-discussion] Style for pad implementation in 'pad' namespace or functions under np.lib
Travis Oliphant travis@continuum...
Mon Apr 2 14:49:09 CDT 2012
On the one hand it is nice to be explicit. On the other hand it is nice to have keyword arguments.
In this case it is very true that pad(a) would not be very clear. Most clear, though, would be: pad(a, width=5, mode='mean'). You could use keyword arguments with None as the default and raise an error if a correct value is not passed in.
On Apr 2, 2012, at 1:14 PM, Tim Cera wrote:
> I think the suggestion is pad(a, 5, mode='mean'), which would be consistent with common numpy signatures. The mode keyword should probably have a default, something commonly used. I'd suggest 'mean', Nathaniel suggests 'zero', I think either would be fine.
> I can't type fast enough. :-) I should say that I can't type faster than Travis since he has already responded....
> Currently that '5' in the example above is the keyword argument 'pad_width' which defaults to 1. So really the only argument then is 'a'? Everything else is keywords? I missed that in the discussion and I am not sure that it is a good idea. In fact as I am typing this I am thinking that we should have pad_width as an argument. I hate to rely on this, because it tends to get overused, but 'Explicit is better than implicit.'
> 'pad(a)' would carry a lot of implicit baggage that would mean it would be very difficult to figure out what was going on if reading someone else's code. Someone unfamiliar with the pad routine must consult the documentation to figure out what 'pad(a)' meant whereas "pad(a, 'mean', 1)", regardless of the order of the arguments, would actually read pretty well.
> I defer to a 'consensus' - whatever that might mean, but I am actually thinking that the input array, mode/method, and the pad_width should be arguments. The order of the arguments - I don't care.
> I realize that this thread is around 26 messages long now, but if everyone who is interested in this could weigh in one more time about this one issue. To minimize discussion on the list, you can add a comment to the pull request at https://github.com/numpy/numpy/pull/242
> Kindest regards,
> Tim
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion@scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
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Trig Vs. Calculus? Which is harder?
calculus involves the concepts of derivatives and integrals of functions. trig functions are one class of functions. so trig is more the study of one class of examples and calculus is an idea.
in practice one applies the idea behind calculus to examples like those found in trig.
thus if you study calculus purely abstractly, it might seem easier than trig, but if you study the examples of calculus, then trig will be a necessary prerequisite to doing calculus in many cases of
practical interest.
I myself learned advanced calculus of banach spaces a la loomis and sternberg before learning trig. a kind of goofy progression. i could prove the graph of a function of bounded variation had measure
zero before i learned to integrate tan(x).
i do not recommend this order of topics. in general, walk first, then run.
but one could learn first the calculus of polynomial functions, before knowing trig.
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[SOLVED] prove that this series to the nth^2 power converges
September 10th 2009, 05:49 PM #1
Senior Member
Feb 2008
[SOLVED] prove that this series to the nth^2 power converges
Prove that
My instinct is to use the root test, and try to show that
But unfortunately I don't know how to show this either.
Are there any tricks I can use on these nth root proofs which might help here?
Note that for each $n\ge1$ is $\left( 1-\frac{1}{n} \right)^{n}\le \frac{1}{e},$ and then $\left( 1-\frac{1}{n} \right)^{n^{2}}\le e^{-n},$ besides the series $\sum_{n\ge1}e^{-n},$ converges by
the integral test, thus the original series does converge.
Of course your idea does work too. In order to prove that $\left( 1-\frac{1}{n} \right)^{n}\to \frac{1}{e}$ as $n\to\infty,$ we have $1-\frac{1}{n}=\frac{n-1}{n}=\frac{1}{\frac{n}{n-1}}=\frac{1}
{1+\frac{1}{n-1}}\implies \left( 1-\frac{1}{n} \right)^{n}=\frac{1}{\left( 1+\frac{1}{n-1} \right)^{n}},$
so it's easy to see that $\left( 1+\frac{1}{n-1} \right)^{n}\to e$ as $n\to\infty,$ and hence, the result.
Thank you very much, but I could still benefit from further assistance.
Once we have $\left( 1-\frac{1}{n} \right)^{n}\le \frac{1}{e}$, the proof is easy for me. However, I am uncertain how you arrived at that inequality.
We have by definition
I have no idea how to prove that this series is increasing, but let's say for a moment that I could. In that case, we have, for each $m=n-1$,
Then algebraically we can show that
Substitute back $m=n-1$ to get
and expand:
But this isn't good enough to get to our goal, which, again, is:
$\left( 1-\frac{1}{n} \right)^{n}\le \frac{1}{e}$.
Apparently I need to learn more about exponents. Any more tricks would be much appreciated.
September 10th 2009, 06:00 PM #2
September 10th 2009, 07:04 PM #3
Senior Member
Feb 2008
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Summary: arXiv:1104.2301v1[math.GR]12Apr2011
Abstract. Geometric semigroup theory is the systematic investigation
of finitely-generated semigroups using the topology and geometry of
their associated automata. In this article we show how a number of
easily-defined expansions on finite semigroups and automata lead to sim-
plifications of the graphs on which the corresponding finite semigroups
act. We show in particular that every finite semigroup can be finitely
expanded so that the expansion acts on a labeled directed graph which
resembles the right Cayley graph of a free Burnside semigroup in many
1. Introduction 2
2. The topology of directed graphs 2
2.1. Directed graphs 3
2.2. Morphisms of directed graphs 6
2.3. Semigroups and automata 7
2.4. Rooted graphs 9
2.5. The unique simple path property 14
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Is the Gouvea-Mazur problem related to symmetric square $L$-functions?
up vote 10 down vote favorite
Here's an idea that I've found appealing but have never been able to get anywhere with.
One way to frame the Gouvea-Mazur question (for lack of a better term, since the original conjecture by the same name has been disproven) is as follows: Let $\pi:\mathcal{E}\to \mathcal{W}$ be the
eigencurve (of some tame level) with its natural projection to weight space. Given a classical point $x\in \mathcal{E}$, what is the radius of the largest disk around $\pi(x)\in \mathcal{W}$ over
which $\pi$ admits a section sending $\pi(x)$ to $x$?
Ramification points of the map $\pi$ provide natural obstructions to the existence of such a section. On the other hand, these ramification points arise naturally as the zeros of a symmetric square
$L$-function on $\mathcal{E}$. This is the the main result of Walter Kim's 2006 thesis written under Robert Coleman (which, as far as I know, never appeared anywhere).
I once had a chat with Barry Mazur in which he (very roughly - these are my words from my recollection several years on) said that this symmetric square $L$ function should be thought of as a section
of $\Omega^1_{\mathcal{E}}$, namely the pullback of the differential $dt/t$ on $\mathcal{W}$ via $\pi$. (Here $t$ is the coordinate on the disk of radius $1$ about $1 \in \mathbb{C}_p$ and we're
using the usual identification of $\mathcal{W}$ with a stack of such disks.)
Clearly $\omega=\pi^*(dt/t)$ has zeros precisely at the ramification points. More generally:
1) Is the "size" of $\omega$ related to the maximal radius of a section to $\pi$?
2) Is this radius then related to the algebraic part of the symmetric square $L$-function?
The first question isn't so well-posed, and may rely on an integral model (formal scheme) giving rise to $\mathcal{E}$.
"should be thought of as a section" - is some version of this literally true? – David Hansen Feb 14 '11 at 21:52
@David: I don't know, but the fact that their zeros coincide is somehow compelling. My recollection is that Mazur was musing about "where $L$-values naturally lie" as objects on the eigencurve,
and that, in the case of the symmetric square $L$-function, the answer seems to be $\Omega^1$. – Ramsey Feb 14 '11 at 22:07
This really confuses me. If you consider some $p$-adic symmetric square $L$-function then its value is a $p$-adic number, so how can it be considered as a differential? I had always thought that
(special values of) $p$-adic $L$-functions were just going to be meromorphic functions on the eigencurve. – Kevin Buzzard Feb 14 '11 at 22:17
To employ a potentially dangerous analogy - a cusp form $f$ of weight two is just a function and it's values are just complex numbers. Nonetheless, there is good reason for thinking of such a
1 thing as a differential form (in this because of the functional equations satisfied by $f$). It wouldn't surprise me if there was something much deeper behind Mazur's comment, but what I took from
the conversation was: since the zeros of $L(Sym^2)$ are exactly the zeros of $\omega$, does this mean that there is some natural sense in which these two objects coincide? – Ramsey Feb 14 '11 at
@Ramsey: I don't buy it! A cusp form of weight 2 is a function on the upper half plane, not a function on a modular curve. The sheaf of differentials on the upper half plane is canonically
trivial; the sheaf of differentials on a modular curve is not. So from that optic it's not surprising at all that a weight 2 modular form is a function---it's just not a function on the right
space :-) – Kevin Buzzard Feb 15 '11 at 7:26
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Browse other questions tagged modular-forms rigid-analytic-geometry l-functions or ask your own question.
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Yet another blog about WPF, Surface, SL, MVVM, NUI....
Calculate the real difference between two angles, keeping the correct sign
1 April 2009
When you build some animations with WPF, Surface or JavaFX you sometimes need to know how evolve an angle. For example, you have the new angle (orientation) of an object and you have store before the
last measure of this orientation : how to calculate the evolution ?
A first solution
"This is simple" will you say, just do this :
double difference = secondAngle - firstAngle;
But this snippet leads to errors. This is because the orientation value given is relative to a certain initial value, and restart to 0° if you pass the 360°. Lets give you an example: if the object
was oriented at 358° and the user turns it of 5° you will obtain a difference of (3-358=) -355° where you actually wants to find 5....
A better solution
A solution I propose is to consider that the methods is called enough frequently that if the object rotate to the left or to the right it has not enough time to rotate more than 180° (half a tour).
Based on this, we consider that the direction of the rotation is given by the "shortest way". If it shorter to turn to the left to go to the new angle, then we select the angle sign which tells we
have turned to the left. It may be easiest to understand by looking atentivly to the image above.
An image is maybe better than word :
We then have this method in C#:
private double calculateDifferenceBetweenAngles(double firstAngle, double secondAngle)
double difference = secondAngle - firstAngle;
while (difference < -180) difference += 360;
while (difference > 180) difference -= 360;
return difference;
A case of use
When do use it ? For example when you build a carousel: the user can click on a specific item and the carousel rotate so it is in front of you. You then need to have the correct angle. I found no
other way to do it.
Does someone have a better way to proceed ?
1. On Wednesday, April 8 2009, 16:48 by LEM
Thanks. it was very useful.
2. On Wednesday, April 8 2009, 17:31 by Jonathan ANTOINE
@LEM : You are welcome !
3. On Thursday, June 18 2009, 16:59 by Dan
Wouldn't if statements make more sense than while loops?
Thanks for the tip.
4. On Thursday, June 18 2009, 17:03 by Jonathan ANTOINE
@[Dan] : The angles can ve really more than 360 and it will then works with this code.
Also I don't think using a while instead of an if cause a lot of perf problems...
5. On Sunday, October 4 2009, 21:21 by Peter
You saved my day
6. On Saturday, November 7 2009, 12:43 by Aubrey
Yep. We use this where I work (and in my homebrew). We call it "Normalize180", and there's a variant called "Normalize360", which does the same thing, but in the range 0f<a<360f.
You could also divide the difference by 180.0f, then take only the fractional part of that (behind the '.' ) and multiply it by 180, making sure to maintain the sign.
float NormalizedAngle = diff / 180.0f;
if(diff > 0){
return (NormalizedAngle - Math.Floor(NormalizedAngle)) * 180.0f;
} else {
return (( NormalizedAngle - ( Math.Floor(NormalizedAngle) + 1.0f ) * 180.0f );//Errr... not sure about this bit - have to deal with negative case. Might be a better way to get the fractional part
than number minus floor(number)
That would probably be less cycles than a while loop, but typical input numbers are unlikely to be high, so your approach is probably fine in most cases.
7. On Saturday, November 28 2009, 22:40 by tk
I was wondering two things: why do you use while instead of if, is that C#? and, perhaps, could you do
double difference = mod( secondAngle - firstAngle + 180, 360 ) - 180;
I suppose it depends on the compiler or deeper knowledge of the computer to know which is faster.
8. On Saturday, November 28 2009, 23:49 by Jonathan ANTOINE
@tk :I simply do it to have the right result : the angle can be more than 360° (suppose you have done 2 turns)....
Your solution may works too...
9. On Friday, January 22 2010, 11:17 by Liz
@tk : That's so elegant - thank you, just what I needed :o)
10. On Monday, March 29 2010, 21:02 by Rich
@tk is superior as the sign can interpret the direction of rotation. Jon's method doesn't work in some cases, say 10 degrees to 350 reports 20 degree turn, not -20 degree...
11. On Tuesday, April 27 2010, 15:55 by René
350 - 10 = 340 > 180 => 340 - 360 = -20
12. On Friday, April 8 2011, 12:09 by Alan
I use:
differenceAngle = (targetAngle - currentAngle) % 360.
this gives you the angle of the target relative to current angle, and is positive for a clockwise rotation.
13. On Sunday, April 24 2011, 16:42 by AronT
A snippet of code taken from my archive in Blitz3D that was used to turn a missile towards the player
It eliminates the need to use mod
Will work with angles of any size
Returns an angle in the range -180 to 180
diffangle = (actualangle - destangle) + 180;
diffangle = (diffangle / 360.0)
diffangle = ((diffangle - Floor( diffangle )) * 360.0) - 180;
!Notes: The Floor command takes the float number and rounds it down to the nearest whole number!
14. On Thursday, April 28 2011, 22:09 by marco polo
try this..
private static double GetPositiveAngle(double radian)
if (radian < 0)
radian = (Math.PI * 2) - Math.Abs(radian);
return radian;
public static double GetAngleDifference(double angle1, double angle2)
var rslt1 = GetPositiveAngle(angle1);
var rslt2 = GetPositiveAngle(angle2);
if (rslt2 < rslt1)
rslt2 += 2 * Math.PI;
return rslt2 - rslt1;
15. On Wednesday, May 4 2011, 04:31 by RNH
If you only want to know the absolute difference then you can use
DEL = PI - ABS(PI - ABS(A - B)) in radians.
DEL = 180 - ABS(180- ABS(A - B)) in degrees.
16. On Thursday, January 12 2012, 13:18 by tigrou
Perfect ! just what i was looking for. Tested and works successfully !
17. On Friday, April 6 2012, 09:02 by JonathanANTOINE@falsemail.com
Cool !
18. On Monday, October 22 2012, 00:06 by AlexL
Thanks, this helped me out big time!
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The Church of Photoshop
is a fine
editor. Like most of 'em, it's got a "
" feature, which
ly (or
ly, if I may
geek out
for a brief moment) lightens or darkens
s in the selected area. The effect is a lot like "
" on a
. It adds
to the signal, so to speak.
Well, it's
. This thing generates
, whatever) patterns of
s. Just like the infinite
monkeys with typewriters
thing, if you beat a series of
random number
s against each other for long enough, damn near anything is liable to fall out --
. If you've got some serious patience.
I've got
running right now. If I just keep on applying "
" to this little
here, I'll eventually see the face of
Our Lord
, or maybe even the
Think of the picture as a number, where each
is a "digit" having a range of values equal to the number of colors available: In a 32 x 32 area, that's
(== 32 * 32) "digits". If that each
could only be black or white, then each "digit" has a range of two possible values. This gets us a number of possible unique arrangements of
s equal to 2 to the
th power, which can be (and should be, if you want to do anything useful with it) viewed as a
-digit number in
base two
We're not limited to two colors, though. We've got 256 colors. Given a 32 x 32
image using
(which is what we're working with here in
), our number is
to the
: The
(or "base") to the power of the number of "digits", as above.
to the
is a lot. (in the neighborhood of 1 followed by 2466 zeroes, I believe).
I'm going to be here for a long time, but you know what,
? I'm ready when you are.
You could express the same number as 2 to the 8192nd power, of course: Eight bits per pixel times 1024 pixels is 8192 bits. In plain English, that's a 8192-digit number in base 2. But if that makes
any sense to you, you know it already.
: Good question. I don't have
Kai's Power Tools
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Do you recognise this variant of the cubes operad?
up vote 6 down vote favorite
In the subject of operads one comes across many quirky variants of more or less the same operad. The cubes operad has many incarnations with various interesting properties. In a recent paper I came
across one such variant. It's so simple I presume other people have come across it before, so it would be nice to have a reference if that's the case. I'll give a sketch of this rather simple
construction, below.
Let me describe the operad, what I think it's good for, and how it relates to other operads. First, I'll set up some notation convention with the cubes operad.
Def'n: (cubes) An increasing affine-linear function $[-1,1] \to [-1,1]$ is a little interval. A product of little intervals $[-1,1]^n \to [-1,1]^n$ is a little $n$-cube. The space $\mathcal C_n(j)$
is the collection of $j$-tuples of little $n$-cubes whose images are required to have disjoint interiors, $\mathcal C_n(0)=\{*\}$ is the empty cube. The collection $\mathcal C_n = \sqcup_{j=0}^\infty
\mathcal C_n(j)$ is the operad of little $n$-cubes, it is a $\Sigma$-operad with structure maps
$$\mathcal C_n(k) \times \left( \mathcal C_n(j_1) \times \cdots \times \mathcal C_n(j_k) \right) \to \mathcal C_n(j_1+\cdots+j_k)$$
defined by
$$(L,J_1,\cdots,J_k) \longmapsto (L_1 \circ J_1, \cdots, L_k \circ J_k)$$
and $\mathcal C_n(j) \times \Sigma_j \to \mathcal C_n(j)$ given by $(L, \sigma) \longmapsto L\circ \sigma$.
Def'n: (overlapping cubes) A collection of $j$ overlapping $n$-cubes is an equivalence class of pairs $(L, \sigma)$ where $L=(L_1,\cdots,L_j)$, each $L_i$ is a little $n$-cube and $\sigma \in \
Sigma_j$. Two collections of $j$ overlapping $n$-cubes $(L,\sigma)$ and $(L',\sigma')$ are taken to be equivalent provided $L = L'$ and whenever the interiors of $L_i$ and $L_k$ intersect $\sigma^
{-1}(i) < \sigma^{-1}(k) \Longleftrightarrow \sigma'^{-1}(i) < \sigma'^{-1}(k)$. Given $j$ overlapping $n$-cubes $(L_1,\cdots,L_j,\sigma)$ say the $i$-th cube $L_i$ is at height $\sigma^{-1}(i)$. $\
sigma(1)$ is the index of the bottom cube, and $\sigma(j)$ is the index of the top cube. Let $\mathcal C_n'(j)$ be the space of all $j$ overlapping $n$-cubes, with the quotient topology induced by
the equivalence relation.
The structure map $$\mathcal C_n'(k) \times \left( \mathcal C_n'(j_1) \times \cdots \times \mathcal C_n'(j_k) \right) \to \mathcal C_n'(j_1 + \cdots + j_k)$$
is defined by
$$\left((L,\sigma), (J_1,\alpha_1), \cdots, (J_k, \alpha_k)\right) \longmapsto ((L_1\circ J_1, \cdots, L_k\circ J_k), \beta)$$
the permutation $\beta$ is given for $1 \leq a \leq k$, $1 \leq b \leq j_a$ by
$$\beta^{-1}\left(\sum_{i<a} j_i + b\right) = \left( \sum_{i<\sigma^{-1}(a)} j_{\sigma(i)}\right) +\alpha^{-1}_a(b)$$
This permutation is obtained by taking the lexicographical order on the set $\{(a,b) : a \in \{1,\cdots,k\}, b \in \{1,\cdots,j_a\}\}$ and then identifying with $\{1, 2, \cdots,j_1+\cdots+j_k\}$ in
the order-preserving way.
== The point ==
So there is a map of operads $\mathcal C_{n+1} \to \mathcal C'_n$ given by sending $(L_1, \cdots, L_j)$ to $(L_1^\pi, \cdots, L_j^\pi, \sigma)$ where we write $L_i = L_i^\pi \times L_i^\nu$ where
$L_i^\pi$ is an $n$-cube and $L_i^\nu$ a $1$-cube. The permutation $\sigma$ is any element $\sigma \in \Sigma_j$ such that $L_{\sigma(j)}^\nu(1) \geq L_{\sigma(j-1)}^\nu(1) \geq \cdots \geq L_{\sigma
Some of the nice things about this operad are:
• (1) it's a multiplicative operad, the inclusion of the associative operad is given by the elements $(Id_{\mathcal [-1,1]^n}, \cdots, Id_{\mathcal [-1,1]^n}, Id_{\{1,\cdots,j\}}) \in \mathcal C'_n
• (2) The map above $\mathcal C_{n+1} \to \mathcal C'_n$ is an equivalence of operads.
• (3) While $\mathcal C_{n+1}$ acts on spaces such as $\Omega^{n+1} X$, $\mathcal C'_n$ does not. $\mathcal C'_n$ acts on spaces of the form $\Omega^n M$ where $M$ is a topological monoid.
The operad of overlapping intervals $\mathcal C'_1$ has a certain affinity to the cactus operad. For example, imagine $[-1/2,1/2]$ as an element of $\mathcal C'_1(1)$ as being represented by $[-1,1]$
with a $1$-cell attached at the points $-1/2$ and $1/2$.
And there are all kinds of variants of this idea -- overlapping discs, or overlapping framed discs, etc. So you can get cyclic multiplicative operads out of these types of constructions.
The criterion for getting the answer "right" is either showing me an occurance of this operad in the literature, or coming up with some convincing argument it's a new construction.
I like this model of the little cubes operad and haven't met it before. Do you think that there's a map from the overlapping intervals operad to a cactus operad? Your remark seems to go half the
way to constructing one. – James Griffin Sep 15 '10 at 14:53
I don't think there's a map but I believe there should be a zig-zag of maps relating the two, essentially staying in the same "circle of ideas" as the above. The intermediate step would be to
compactify the operad of overlapping intervals, to allow for infinitesimal intervals. I'm not sure which precise compactification you'd want to use, but the cactus operad would be a sub-operad of a
suitable compactified overlapping intervals operad. My understanding is Paolo Salvatore had an MSc student who found a quite different map to the cactus operad, using a leaf-space of a foliation
idea. – Ryan Budney Sep 15 '10 at 16:41
add comment
1 Answer
active oldest votes
I just now noticed this question.
Your operad is a suboperad of the "surgery cylinder" operad described in arxiv 1009.5025 (more recent version available here). See Section 8 and figures therein. In the notation of that
paper, your operad corresponds to case where all the manifolds $M_i$ and $N_i$ are $n$-cubes and all the homeomorphisms $f_i$ are the identity.
up vote 4
down vote The surgery cylinder operad can be thought of as describing a sequence of (generalized) surgeries on an initial manifold $M_0$, yielding a final manifold $N_0$. At the $i$-th stage ($1\
accepted le i \le k$) we remove a codimension-0 submanifold $M_i$ replace it with $N_i$, where $m_i$ and $N_i$ have the same boundary. To get your overlapping $n$-cubes operad, let $M_0$ be the
"big" $n$-cube and $M_i = N_i$ be the $i$-th little $n$-cube. In other words we have a sequence of pointless surgeries in which little $n$-cubes are removed and replaced with identical
copies of themselves.
@Kevin: thanks. I'll take a closer look at your paper. On first glance it looks like your "surgery cylinder operad" may be related to my "splicing operad" front.math.ucdavis.edu/
1004.3908 – Ryan Budney Apr 15 '11 at 14:29
@Ryan: Thanks for pointing out your paper to me. I agree there should be some relation between the two operads. Once I know more precisely what the relation is, I'll add a remark to
that effect to my (and Scott Morrison's) paper. – Kevin Walker Apr 15 '11 at 15:25
add comment
Not the answer you're looking for? Browse other questions tagged operads or ask your own question.
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Electrodynamics/Vector Calculus Review
From Wikibooks, open books for an open world
This page is going to review some of the necessary background information in physics and vector calculus. In this page, we will make extensive use of an analogy between vector fields and the flow of
water so that you will gain intuitive understanding of the material.
The Del Operator[edit]
The del operator, ∇ is defined as follows:
$abla = \hat{x}\frac{\partial}{\partial x } + \hat{y}\frac{\partial }{\partial y } + \hat{z}\frac{\partial }{\partial z }$
This operator, while confusing at first, is the method by which vectors and scalars can be differentiated.
The Gradient[edit]
When ∇ operates on a scalar field, like so:
$abla\Phi = \hat{x}\frac{\partial\Phi}{\partial x } + \hat{y}\frac{\partial\Phi }{\partial y } + \hat{z}\frac{\partial\Phi }{\partial z }$
it simultaneously differentiates the scalar by all 3 axes (x, y, z). The result is called the "Gradient" of the scalar. The gradient is a vector that points in the direction in which the original
scalar field is changing most rapidly (has the largest derivative).
In addition, $abla\Phi \cdot \hat{n}$ gives the rate of change of $\Phi$ in the direction of $\hat{n}$. Thus, the perpendicular lines to the gradient form an equipotential surface, or a surface where
$\Phi$ are all equal.
Example 1.1
The function $F(x,y) = x^2 - y^2$ and $abla F$ are demonstrated in the associated diagram. The key concepts here are that the vectors of the gradient point towards the higher magnitude of $F(x,y)$
and that the vector represents the rate of change between the origin and head of the this vector.
If $F(x,y)$ was a uniformly rigid surface and a perfect sphere was placed exactly at $F(2,0)$ it would move towards $F(-2,0)$ and eventually settle at $F(0,0)$. In this case, if y is ever non-zero
then the ball will eventually fall off the surface.
The Divergence[edit]
The ∇ operator can be loosely treated as a "vector" whose components are the partial differential operators. If we operate on a vector field as a "dot product", we obtain:
$abla\cdot A = \frac{\partial A_x}{\partial x } + \frac{\partial A_y }{\partial y } + \frac{\partial A_z }{\partial z }$
This is called the "Divergence" of the vector field. It measures how much the vector "diverges" from a single point. It measures the "sources" and "sinks" of the vector field. Imagine the velocity
vector field of a pool of water. The faucets are places of high positive divergence, because it is the source of the water velocity field, and the sinks (drains) are places of high negative
divergence, because this is where all the water is converging.
The Curl[edit]
If we cross ∇ onto a vector field, we obtain another important operator:
$abla\times A = \hat{x}(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}) + \hat{y} (\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}) + \hat{z} (\frac{\partial A_y}
{\partial x}-\frac{\partial A_x}{\partial y})$
The resulting vector is the "Curl" of the original vector. It measures the "curling" or the "rotation" of the vector field at a single point. Thus, going back to the pool analogy, a whirlpool would
be a place with a large curl. In a steady flow, the curl is 0, since the field doesn't want to curl around that point.
The Laplacian[edit]
The gradient ∇Φ introduced above is a vector field. What happens if we take its divergence?
$abla\cdot abla\Phi = abla^2\Phi = \frac{\partial^2 \Phi}{\partial x^2} + \frac{\partial^2 \Phi}{\partial y^2} + \frac{\partial^2 \Phi}{\partial z^2}$
This important operator is known as the "Laplacian". The Laplacian is also defined for vector fields:
$abla^2 A = \hat{x}abla^2 A_x + \hat{y}abla^2 A_y + \hat{z} abla^2 A_z$
The Divergence of the Curl[edit]
One might also expect to obtain an important operator by taking the divergence of a curl:
$abla\cdot (abla \times A) = \frac{\partial}{\partial x}(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}) + \frac{\partial}{\partial y}(\frac{\partial A_x}{\partial z}-\frac{\
partial A_z}{\partial x}) + \frac{\partial}{\partial z}(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}) = 0$
While zero is certainly an important concept, it does not provide us with a useful operator. This identity, however, is interesting in its own right. It turns out that every vector field that is
divergence-free is the curl of another vector field.
The Curl of the Gradient[edit]
$abla \times abla\Phi = \hat{x}(\frac{\partial^2 \Phi}{\partial z \partial y} - \frac{\partial^2 \Phi}{\partial y \partial z}) + \hat{y}(\frac{\partial^2 \Phi}{\partial z \partial x} - \frac{\
partial^2 \Phi}{\partial x \partial z}) + \hat{z}(\frac{\partial^2 \Phi}{\partial y \partial x} - \frac{\partial^2 \Phi}{\partial x \partial y}) = 0$
It also turns out that every vector field that has no curl is the gradient of a scalar field.
Vector Fields[edit]
Vector fields are 3 dimensional volumes, for which every point within that volume can be assigned a vector magnitude, based on some given rule. Gravity is one example of a vector field, where every
point within a gravitational field is being pulled with some force magnitude towards the center. A vector field is denoted by a 3-dimensional function, such as A(x, y, z). The value of the function
for each triplet is the magnitude of the vector field at that point.
In speaking of vector fields, we will discuss the notion of flux in general, and electric flux specifically. We can define the flux of a given vector field G(x, y, z), through an infinitesimal area
dA, which has a normal vector n:
$\operatorname{Flux}(dA, n) = G \cdot n dA$
Which we read as "The flux passing through dA, in the direction of n".
The concept of flux originally came from hydrodynamics. The flux passing through a small surface is the amount of liquid that flows through it. If the velocity field is big, then the flux would
naturally turn out to be big. Also, if the surface is parallel to the velocity field, then the flux is 0, because no water is passing through the area. It turns out flux is a very useful concept in
Electrodynamics also.
If we integrate this equation with respect to dA, we get the following:
$\operatorname{Flux}(A, n) = \int_A v \cdot dA$
We can also show (although the derivation can be long), that the flux traveling into or out of a given vector field, G, can be given by the divergance of the vector field:
$\operatorname{Net Flux} = abla \cdot G$
Let's say that we have an arbitrary volume, V, in a vector field, G, bounded by a surface, S, with surface-area, A. Gauss' Theorem states that the flux flowing into this volume is equal to the amount
of flux flowing through the surface, S.
$\int_{V} (abla \cdot G) dV = \int_{S, V} (v \cdot n) dA$
This formula is intuitively true because as we seen before, the divergence of a field is how much the field spreads from that point. If we add the spreading of every point inside a volume, that
should be the amount that is leaving the volume through the closed surface. The flux through any closed surface of a divergence free field is 0.
Line Integration[edit]
Suppose we have a path $\gamma$, composed of small length elements $ds$. The line integral of a vector field A is how much the field lies along the path. Formally, it is given by:
$\int_{\gamma} A \cdot ds$
Suppose that A is a curl-free field. As seen above, this means it could be written as the gradient of a certain field. Say that $A=abla \Phi$. It turns out that the line integral of A along any path
connecting two points are the same. Moreover,
$\int_{\gamma} A \cdot ds =\Phi(b)-\Phi(a)$
where a and b are the start and end points of the path. Also, the line integral along a closed loop of such a field is always 0. A curl-free field is called conservative. The reason for this
terminology came from mechanics: In mechanics, if you have a force field in space that is curl-free, you can always define a potential energy function, so that the work done in moving an object from
a to b is the difference in potential energy. In this case, mechanical energy is conserved.
As we will see, the electrostatic field is a conservative field, whereas the magnetic field and the general electric field are not.
In general, however, the line integral of a vector field along a closed loop is nonzero. It turns out, however, that:
$\int_{\gamma} F \cdot ds=\int_{A} (abla \times F) \cdot dA$. In other words, the line integral of a vector field is equal to the flux of the curl of the field through the loop. Through which
surface are we finding the flux? Any surface! Any surface enclosed by the loop will suffice.
An intuitive understanding of this theorem can come as follows: the line integral of a closed loop is like how much the field wraps around the loop. However, the curl at a point gives how much the
field rotates around it. Thus, the total integral over the surface gives the total "wrapping" around the loop.
Divergence and Curl[edit]
We have shown that the divergence of an arbitrary vector A is given by:
$\operatorname{Divergence}(A) = abla \cdot A$
and likewise, we define an operator called Curl that acts on a vector field and is defined as such:
$\operatorname{Curl}(A) = abla \times A$
We will be using divergence and curl throughout the rest of the chapters on electromagnetism.
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Lever calculations
My problem (trying to help an ay last keen son)-a lever 5m long, a weight of 300kg, a force of 100kg.
The question says describe?
I am trying to work out the distance from the end of lever to fulcrum
or in other words where the fulcrum should lie.
tried formula which has gone to course with son but it gives L over little l = force over effort?
Can you assist please?
I feel there must be some better formula.
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only answers!
1 (true or false) Whenever the IRR on a project equals that project’s required
only answers!
1 (true or false) Whenever the IRR on a project equals that project’s required rate of return, the NPV equals zero.
2 If the NPV of a project is positive, then the project’s IRR ____________ the required rate of return.
a. must be less than
b. must be greater than
c. could be greater or less than
d. cannot be determined without actual cash flows
3 Which of the following capital-budgeting techniques assume the same reinvestment rate of cash flows?
a. MIRR
b. NPV
c. IRR
d. Both a & b
e. All of the above
4 Payback period:
a. ignores the time value of money.
b. deals with cash flows rather than accounting profits.
c. measures how quickly the project will return its original investment.
d. does all of the above.
5 As the cost of capital is increased, the:
1. IRR remains constant.
2. payback period remains the same.
3. discounted payback period increases.
4. both b and c.
5. all of the above.
6 Which of the following is the correct equation to solve for the NPV of the project that has an initial outlay of $30,000, followed by three years of $20,000 in incremental cash inflow? Assume a
discount rate of 10%.
a. NPV = -$30,000 + $20,000(1.10)1 + $20,000(1.10)2 + $20,000(1.10)3
b. NPV = -$30,000 + $20,000(1.10)-1 + $20,000(1.10)-2 + $20,000(1.10)-3
c. NPV = -$30,000 + $20,000/(1.01).10 + $20,000/(1.02).10 + $20,000/(1.03).10
d. NPV = -$30,000 + $20,000/(1.1).10+ $20,000(1.2).10 + $20,000(1.3).10
7 Why does the NPV method of evaluating an investment proposal require that the cash inflows of a project be discounted to the present?
a. It is the only way to arrive at the correct amount to divide into the cost in order to determine the rate of return.
b. The IRS requires it.
c. This enables the analyst to determine the amount of the investment outlay.
d. It provides a measurement of the value of an investment proposal in terms of today’s dollars.
8 Kannan Enterprise has a project with an initial outlay of $40,000, followed by three years of annual incremental cash flows of $35,000. The terminal cash flow of the project is $10,000. Assuming a
discount rate of 10%, which of the following is the correct equation to solve for the IRR of the project?
a. $40,000 = $35,000(1.12)1 + $35,000(1.12)2 + $45,000(1.12)3
b. $40,000 = $35,000(1 + IRR)1 + $35,000(1+IRR)2 + $45,000(1+IRR)3
c. $40,000 = $35,000/(1.12)IRR + $35,000/(1.12)IRR + $45,000/(1.12)IRR
d. $40,000 = $35,000(1+IRR)-1 + $35,000(1.IRR)-2 + $45,000(1+IRR)-3
9 Mayhem Mines, Inc. is analyzing a project that requires an initial investment of $50,000, followed by cash inflows of $15,000 in Year 1, $60,000 in Year 2, and $75,000 in Year 3. The cost of
capital is 10%. Calculate the profitability index of the project.
a. 2.14
b. 2.26
c. 2.39
d. 2.47
10 Mayhem Mines, Inc. is analyzing a project that has a profitability index of 2.5. Given the following cash flows for the project, calculate the present value of inflows for the project.
Year Cash Flow
0 ($100,000)
1 $120,000
2 $130,000
3 $200,000
a. $250,000
b. $350,000
c. $450,000
d. $550,000
11 So long as money has a time value, the discounted payback will always be greater than the:
a. MIRR.
b. IRR.
c. NPV.
d. payback.
12 (true or false) The capital rationing problem can be correctly solved by ranking projects according to the profitability index.
13 (true or false) Working capital for a project includes investment in fixed assets.
14 (true or false) If the firm decides to impose a capital constraint on investment projects, the appropriate decision criterion is to select the set of projects that has the highest net present
value subject to the capital constraint.
15 (true or false) Sales captured from the firm’s competitors can be relevant to the capital-budgeting decision.
16 (true or false) NPV and IRR can provide ranking inconsistencies when projects have unequal lives.
Use the following information to answer questions 17-19. Delta Inc. is considering the purchase of a new machine which is expected to increase sales by $10,000 in addition to increasing
non-depreciation expenses by $3,000 annually. Due to the sales increase, Delta expects its working capital to increase $1,000 during the life of the project. Delta will depreciate the machine using
the straight-line method over the project’s four year life to a salvage value of zero. The machine’s purchase price is $20,000, and installation fee is $1,000. The firm has a marginal tax rate of 34
percent, and its required rate of return is 12 percent.
17 The machine’s initial cash outflow is:
a. $20,000.
b. $22,000.
c. $27,000.
d. $23,000.
18 The machine’s incremental after-tax cash inflow is:
a. $6,405.
b. $7,980.
c. $8,620.
d. $5,980.
19 The machine’s after-tax incremental cash flow in year four is:
a. $6,980.
b. $5,980.
c. $7,405.
d. $8,620.
20 Depreciation expenses affect tax-related cash flows by:
a. increasing taxable income, thus increasing taxes.
b. decreasing taxable income, thus reducing taxes.
c. decreasing taxable income, but not altering cash flows since depreciation is not a cash expense.
d. all of the above.
e. none of the above.
21 When determining the initial cash outlay of an asset that is being replaced, which of the following should be included?
a. Interest expense that is directly related to the financing of a project
b. The pre-tax sales proceeds of the asset being sold
c. Market study expenses that were incurred in order to decide if the firm should accept a project
d. Principal payments that are directly related to the financing of a project
e. The sales proceeds of the asset being sold, net of any income taxes related to the sale
22 Which of the following should be considered in the estimation of termination cash flows?
a. Cash generated from the sale of a project
b. Recovery of net working capital
c. Income taxes associated with the sale of a project
d. All of the above
23 When evaluating Capital Budgeting decisions, which of the following items should not be included in the construction of cash flow projections for purposes of analysis?
a. Net salvage value
b. Land and building expenses
c. Changes in net working capital requirements
d. Shipping and installation costs
e. All of the above should be included.
24 (true or false) No adjustment is made in the cost of preferred stock for taxes since preferred stock dividends are not tax-deductible.
25 The investor’s required rate of return differs from the firm’s cost of capital due to the:
a. firm’s beta.
b. tax deductibility of interest.
c. CAPM.
d. time value of money.
26 Bender and Co. is issuing a $1,000 par value bond that pays 9% interest annually. Investors are expected to pay $918 for the 10-year bond. Bender will have to pay $33 per bond in flotation costs.
What is the after tax cost of debt if the firm is in the 34% tax bracket?
a. 7.23%
b. 9.01%
c. 9.23%
d. 11.95%
27 The expected dividend is $2.50 for a share of stock priced at $25. What is the cost of retained earnings if the long-term growth in dividends is projected to be 8%?
a. 10%
b. 8%
c. 25%
d. 18%
28 (true or false) As the tax rate increases, the weighted average cost of capital decreases.
29 (true or false) Using a firm’s overall cost of capital to evaluate divisional projects can lead to the rejection of good investments in low-risk divisions and the acceptance of poor projects in
high-risk divisions.
30 (true or false) The data a firm used to calculate its cost of capital is appropriate for varying levels of debt and equity in the firm’s capital structure.
31 (true or false) The cost of capital for common stock is higher than that for bond and preferred stock.
32 Which of the following is considered a problem in using the dividend-growth model to estimate the cost of equity?
a. Estimating the expected growth rate of future dividends
b. Determining the expectations of the minds of the investors
1. Using subjectivity in estimating the risk premium
2. Both a and b
e. All of the above are correct
Use the following information to answer questions 33-35. Sigma Corp. has a target financing mix of 30% debt and 70% common equity. The before-tax cost of Sigma’s debt is 10%. Sigma estimates its cost
of internal equity at 14% and its cost of new common equity at 17%. Sigma’s marginal tax rate is 34%, and it expects to have $2,100,000 of profit available for reinvestment in the firm.
33 If Sigma can meet all of its financing needs with debt and internally generated equity, then Sigma’s weighted cost of capital is:
a. 11.78%.
b. 12.38%.
c. 13.88%.
d. 14.28%.
34 What is the total dollar amount of new investment that Sigma can support without issuing new common equity?
a. $1,500,000
b. $2,000,000
c. $2,500,000
d. $3,000,000
35 If Sigma cannot meet all of its financing needs with debt and internally generated equity, then Sigma’s weighted cost of capital is:
a. 11.78%.
b. 12.38%.
c. 13.88%.
d. 14.28%
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Indirect Measurement - Using Similar Triangles (with videos, worksheets, games & activities)
Indirect Measurement - Using Similar Triangles
Videos, worksheets, stories and songs to help Grade 8 students learn about indirect measurement (using similar triangles).
Indirect Measurements
Apply concepts of similar figures to solve indirect measurement problems
Intermediate Concepts of Indirect Measurement
Learn to use concepts of proportions to solve problems in indirect measurement
Advanced Concepts of Indirect Measurement
Learn to use concepts of proportions to solve problems in indirect measurement
Indirect Measurement :
Indirect measurement is a method of using proportions to find an unknown length or distance in similar figures.
Two common ways to achieveindirect measurement involve
(1) using a mirror on the ground and
(2) using shadow lengths and find an object's height.
Method 1 measures the person's height and the distances between the person, mirror, and object.
Method 2 measures shadows and the person's height.
Custom Search
We welcome your feedback, comments and questions about this site - please submit your feedback via our Feedback page.
Custom Search
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SUMIF Function
NOTICE. While work is in progress, much of this topic is incomplete. You can get more information here. Thanks for your patience.--Charley Kyd
Math and trigonometry
SUMIF Function
Adds the cells in a row or column, as specified by one criteria
by Charley Kyd, MBA
Microsoft Excel MVP
SUMIF summarizes data in a range using one criteria.
SUMIF(range, criteria, sum_range)
• range Required. The range of cells for the criteria to evaluate.
• criteria Required. The criteria in the form of a number (like 50), text (like "sales"), or expression as text (like ">=99"). Text must be enclosed in quotation marks (like: "sales"). You also can
use the wildcard characters "?" (to match a single character) and "*" (to match any number of characters).
• sum_range Optional. The range of cells to add. If this argument is omitted, SUMIF adds values in range.
Applies To
Excel 2003 and above
Keep in mind that SUMIF, which allows one criteria, and SUMIFS, which allows more than one criteria, arrange their arguments in a different sequence. For this reason, it's often more conventient to
use the SUMIFS function for even one criteria.
If you're a subscriber to my Excel for Business newsletter, you can download this example workbook here. Otherwise, you can subscribe here for free, and then download this workbook.
Example 1:
SUMIF returns the sum of all values that are greater than 10.
Example 2:
Because no sum_range is specified, SUMIF must sum "Hats". But "Hats" isn't numeric, so its sum is zero.
Example 3:
SUMIF returns the total for Hats in the Sum_Range.
Example 4:
The case of the Criteria doesn't matter. SUMIF still returns the total for Hats in the Sum_Range.
Example 5:
SUMIF returns the sum of the Sum_Range values for Range values that meet the Criteria, whether they're entered as numbers or text.
Example 6:
SUMIF returns the sum of all numeric Codes greater than 75. Here, the Range item "88" is ignored because it's entered as text and has no numeric value.
Example 7:
SUMIF returns the sum for all Range items starting with "H" and ending with "s".
Example 8:
SUMIF returns the sum of Sum_Range values for Range items like t?es, where "?" can be any character.
Example 9:
SUMIF converts the date expressed as text in the Criteria cell to a date serial number, then returns the proper value.
Example 10:
Here, the Criteria cell isn't empty. Instead, it contains a single-quote ('). It also could contain a null string entered by formula (=""). In either case, SUMIF returns the sum of the Sum_Range
values for Range cells that are empty.
Example 11:
SUMIF adjusts the Sum_Range to be in sync with the Range. Here, when the Sum_Range is only one cell, SUMIF expands it apply to the entire Range column.
Example 12:
SUMIF adjusts the Sum_Range to be in sync with the Range. Here, when the Range is only one cell, SUMIF redues the Sum_Range so that it corresponds with only that one Range cell.
Example 13:
SUMIF adjusts the Sum_Range to be in sync with the Range. When the Range is one column, SUMIF contracts the Sum_Range to be the same.
Other Help
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Physics Forums - View Single Post - Pseudo random number algorithm
I'm trying to figure out this little problem. I'm trying to figure out which algorithm is being used.I know the set of numbers is being used to generate the 5 digit number. I believe a Pseudo random
number generator algorithm is being used. I know the set of number are being used as the seed, not sure if a,b & c are treated as separate variables or they are concatenated. a,b &c are the input and
the 5digit number is the output
a:347 b:851 c:0381 = 86013
a:347 b:547 c:5840 = 54552
a:347 b:258 c:8744 = 49785
a:347 b:852 c:1707 = 39888
Let me know if anyone has any Ideas.
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Universal One Way Hash Functions and Their Cryptographic Applications
- Journal of Cryptology , 1993
"... We show very efficient constructions for a pseudo-random generator and for a universal one-way hash function based on the intractability of the subset sum problem for certain dimensions.
(Pseudo-random generators can be used for private key encryption and universal one-way hash functions for sign ..."
Cited by 78 (8 self)
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We show very efficient constructions for a pseudo-random generator and for a universal one-way hash function based on the intractability of the subset sum problem for certain dimensions.
(Pseudo-random generators can be used for private key encryption and universal one-way hash functions for signature schemes). The increase in efficiency in our construction is due to the fact that
many bits can be generated/hashed with one application of the assumed one-way function. All our construction can be implemented in NC using an optimal number of processors. Part of this work done
while both authors were at UC Berkeley and part when the second author was at the IBM Almaden Research Center. Research supported by NSF grant CCR 88 - 13632. A preliminary version of this paper
appeared in Proc. of the 30th Symp. on Foundations of Computer Science, 1989. 1 Introduction Many cryptosystems are based on the intractability of such number theoretic problems such as factoring and
discrete logarit...
- Journal of Cryptology , 1994
"... A signature scheme is existentially unforgeable if, given any polynomial (in the security parameter) number of pairs (m 1 ; S(m 1 )); (m 2 ; S(m 2 )); : : : (m k ; S(m k )) where S(m) denotes
the signature on the message m, it is computationally infeasible to generate a pair (m k+1 ; S(m k+1 )) fo ..."
Cited by 45 (5 self)
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A signature scheme is existentially unforgeable if, given any polynomial (in the security parameter) number of pairs (m 1 ; S(m 1 )); (m 2 ; S(m 2 )); : : : (m k ; S(m k )) where S(m) denotes the
signature on the message m, it is computationally infeasible to generate a pair (m k+1 ; S(m k+1 )) for any message m k+1 = 2 fm 1 ; : : : m k g. We present an existentially unforgeable signature
scheme that for a reasonable setting of parameters requires at most 6 times the amount of time needed to generate a signature using "plain" RSA (which is not existentially unforgeable). We point out
applications where our scheme is desirable. Preliminary version appeared in Crypto'94 y IBM Research Division, Almaden Research Center, 650 Harry Road, San Jose, CA 95120. Research supported by a BSF
Grant 32-00032-1. E-mail: dwork@almaden.ibm.com. z Incumbent of the Morris and Rose Goldman Career Development Chair, Dept. of Applied Mathematics and Computer Science, Weizmann Institute of Science,
- CS-R9529, Computer Science, Dept. of Algorithms and Architecture, CWI , 1995
"... Even, Goldreich and Micali showed at Crypto'89 that the existence of signature schemes secure against known message attacks implies the existence of schemes secure against adaptively chosen
message attacks. Unfortunately, this transformation leads to a rather impractical scheme. We exhibit a simil ..."
Cited by 3 (1 self)
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Even, Goldreich and Micali showed at Crypto'89 that the existence of signature schemes secure against known message attacks implies the existence of schemes secure against adaptively chosen message
attacks. Unfortunately, this transformation leads to a rather impractical scheme. We exhibit a similar security amplification, which takes the given scheme to a new signature scheme that is not even
existentially forgeable under adaptively chosen message attacks. Additionally, however, our transformation will be practical: The complexity of the resulting scheme is twice that of the original
scheme. The principles of both transformations carry over to block encryption systems. It is shown how they can be used to convert a block encryption system secure against known plaintext attacks to
a system secure against chosen plaintext attacks. For both schemes it is shown that if the transformed scheme can be broken given a number, T , of encryptions of adaptively chosen plaintexts, then
, 1994
"... We outline constructions for both pseudo-random generators and one-way hash functions. These constructions are based on the exact TSP (XTSP), a special variant of the well known traveling
salesperson problem. We prove that these constructions are secure if the XTSP is infeasible. Our constructions a ..."
Cited by 2 (1 self)
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We outline constructions for both pseudo-random generators and one-way hash functions. These constructions are based on the exact TSP (XTSP), a special variant of the well known traveling salesperson
problem. We prove that these constructions are secure if the XTSP is infeasible. Our constructions are easy to implement, appear to be fast, but require a large amount of memory.
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Snow Walkers Puzzle Description
Dennis E. Shasha, Courant Institute of Mathematical Sciences,
New York University.
Grid City is a small planned city laid out at a completely regular 6 by 5 grid
with streets going north-south and east-west.
There is one building per city block.
Grid occasionally
suffers major snow storms.
The Grid City Snow Clearing Department (GridClear)
wishes to make it possible to reach each
city block but the effort to move the snow is so great that
they resolve to make it possible to go from any block to any other
through paved paths.
This means that the path itself may snake through the city
and may even loop, but the plows won't double-back on a paved road
for fear of running over the many pedestrians.
The head of GridClear consults you to help plan the path.
You are told the path must start
somewhere on the outside boundary of the grid, but you
can choose where,
because Grid City has many garages available.
The goal is to ensure that for any two blocks that neighbor
one another north-south or east-west, a resident will
need to travel over only a few streets (or through a few buildings)
along the cleared path.
The "score" of a path is the worst case, i.e. the largest number of
such streets/buildings.
Crossing a plowed intersection costs nothing, but it is impossible
to cross an unplowed intersection or street.
You may assume that each building block has an entrance at every corner
and that walking through a building to any other corner (even
to the diagonally opposite one) costs
the same as walking one city block along a plowed street.
You accept this simplication, because there is no ice inside the buildings.
Note: In the warm-up and questions to follow, you may find it helpful
to look at the very nice applet of Arefin Huq with whom I have worked
on this puzzle.
All screenshots are from that applet.
Suppose the city were smaller.
Given the 3 by 4 city grid of
try to find a route requiring plowing only five streets,
so pedestrians can walk from any block to a neighbor across
at most two streets.
Solution to warm-up:
As the figure
shows, for most blocks, a pedestrian can walk from a corner
across the street to a corner of the neighboring building.
The two blocks marked with 2 require walking through two buildings
to travel to one another. For example, to go from (1,0) to (2,0),
go to the northeast corner of (1,0), cross to (1,1), then go diagonally
to the southeast corner of (1,1) and then cross the plowed street
to the northeast corner of (2,1). From there go to the southwest
corner of (2,1) and then into the block (2,2).
End of Warm-Up.
Recall that our city is 6 by 5 as illustrated in
It's still the case that every block has an entrance at every corner
and you must begin plowing from the outside of the grid.
1. Assuming you have just one plow, can you arrange
a path of 15 or fewer streets
and guarantee that
a pedestrian on any
block can
reach any east-west or north-south neighbor by
walking at most eight blocks?
What does your path look like?
2. Still with just one plow,
what is the minimum length path you can design that will guarantee
that a pedestrian can walk from any block to its neighbor
across a cleared intersection (cost of 0)?
A neighboring city has just given you another plow.
You can therefore use two but both must start from the outside of the grid
and neither can drive on a city block already plowed by the other
(though it can cross an intersection).
3. Can you figure out a way to use two plows, such that each plows
nine streets, in such a way that
that a pedestrian can walk from any block to its neighbor
across a cleared intersection (cost of 0)?
1. As you can see in the solution
plowing just 14 streets is enough and the maximum number of blocks
traveled is just 6.
2. The solution of
requires plowing 18 streets.
That's the best I know of.
3. In the solution of
each vehicle plows nine streets
of Grid City.
All these solutions and the applet are the work of Arefin Huq.
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"fields" and relativity - a broad question
... being a general enquiry into the uses and meanings of the word "field(s)" in relation to Einstein's theories of Special Relativity and General Relativity.
First, the
narrow technical meanings
QFT ('quantum field theory') involves 'fields'. In what sense, if any, does GR (General Relativity) also involve 'fields'. By 'involve' here I mean both the textbook explication of the core equations
and relations(hips), as well as the underlying mathematical structure (to as many levels as you wish). To the extent that they both involve 'fields', how similar are these 'fields'?
Next, the
more general
(but still narrow) meanings.
In 'textbook' material - such as
from Eric Weisstein's World of Physics (I typed 'gravitational field' into Google and chose the first hit that looked 'textbooky') - the concept of a 'gravitational field' seems alive and flourishing
(Google tells me that there are >4 million hits to my simply enquiry; an eyeball estimate of the first half dozen webpages of hits suggests that many of these are from textbook-style webpages
(ignoring crackpot sites, of course)). Do readers of this post have a feel for the most common ways that this expression/concept/term ("gravitational field") is tied to GR - in terms of both the
textbook approach/explication, and the core aspects of the theories (which of course include Newtonian gravity)?
the popsci/folk/general meanings
By 'popsci' here I mean popular science writing on the topic of gravitation and relativity; specifically, that which seeks to explain the ideas, concepts and theories
without using any equations or math
. By 'folk/general' I mean use of the terms outside any of the environments described above, such as on Star Trek, in computer games, literary criticism, etc. This is, of course, a vastly bigger
field than all the above combined, yet it is the one in which more Joe Sixpacks and Joan Chardonnays will encounter 'gravitational field'.
What opinions do readers of this post have concerning the ranges of meanings that are to be found here?
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they are shaking hands together
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Azhar Shahid (azharshahid01@gmail.com)
New Delhi
Time: 00:01:16
Placed User Comments (More)
rajesh dolai
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14 Days ago
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An Awesome site for TCS.
Cleared the aptitude.
1 Month ago
I successfully cleared TCS aptitude test held on 8th march 2014.Thanks a lot m4maths.com
plz guide for the technical round.
mounika devi mamidibathula
1 Month ago
got placed in IBM..
this site is very useful, many questions repeated..
thanks alot to m4maths.com
Anisha Lakhmani
1 Month ago
I got placed at infosys.......thanx to m4maths.com.......a awesum site......
Anisha Lakhmani
1 Month ago
I got placed at infosys.......thanx to m4maths.com.......a awesum site......
Kusuma Saddala
2 Months ago
Thanks to m4maths, i have place at IBM on feb 8th of this month
2 Months ago
thanks to m4 maths because of this i clear csc written test
mahima srivastava
2 Months ago
Placed at IBM. Thanks to m4maths. This site is really very helpful. 95% questions were from this site.
Surya Narayana K
2 Months ago
I successfully cleared TCS aptitude test.Thanks a lot m4maths.com.
Surya Narayana K
2 Months ago
I successfully cleared TCS aptitude test.Thanks a lot m4maths.com.
prashant gaurav
2 Months ago
Got Placed In Infosys...
Thanks of m4maths....
3 Months ago
iam not placed in TCS...........bt still m4maths is a good site.
4 Months ago
Thanx to m4 maths, because of that i able to crack aptitude test and now i am a part of TCS. This site is best for the preparation of placement papers.Thanks a lotttttt............
4 Months ago
THANKS a lot m4maths. Me and my 2 other roomies cleared the tcs aptitude with the help of this site.Some of the questions in apti are exactly same which i answered without even reading the whole
question completely.. gr8 work m4maths.. keep it up.
5 Months ago
m4maths is one of the main reason I cleared TCS aptitude. In TCS few questions will be repeated from previous year aptis and few questions will be repeated from the latest campus drives that happened
in various other colleges. So to crack TCS apti its enough to learn some basic concepts from famous apti books and follow all the TCS questions posted in m4maths. This is not only for TCS but for all
other companies too. According to me m4maths is best site for clearing apti. Kuddos to the creator of m4maths :)
5 Months ago
THANKS A LOT TO M4MATHS.due to m4maths today i am the part of TCS now.got offer letter now.
5 Months ago
Hai friends, I got placed in L&T INFOTECH and i m visiting this website for the past 4 months.Solving placemetn puzzles from this website helped me a lot and 1000000000000s of thanks to this
website.this website also encouraged me to solve puzzles.follw the updates to clear maths aps ,its very easy yar, surely v can crack it if v follow this website.
6 Months ago
2 days before i cleared written test just because of m4maths.com.thanks a lot for this community.
6 Months ago
thanks for m4maths!!! bcz of which i cleared apti of infosys today.
6 Months ago
Today my written test of TCS was completed.I answered many of the questions without reading entire question.Because i am one of the member in the m4maths.
No words to praise m4maths.so i simply said thanks a lot.
7 Months ago
I am very grateful to m4maths. It is a great site i have accidentally logged on when i was searching for an answer for a tricky maths puzzle. It heped me greatly and i am very proud to say that I
have cracked the written test of tech-mahindra with the help of this site. Thankyou sooo much to the admins of this site and also to all members who solve any tricky puzzle very easily making people
like us to be successful. Thanks a lotttt
Abhishek Ranjan
7 Months ago
me & my rooom-mate have practiced alot frm dis site TO QUALIFY TCS written test.both of us got placed in TCS :)
do practice n u'll surely succeed :)
Sandhya Pallapu
1 year ago
Hai friends! this site is very helpful....i prepared for TCS campus placements from this site...and today I m proud to say that I m part of TCS family now.....dis site helped me a lot in achieving
this...thanks to M4MATHS!
vivek singh
2 years ago
I cracked my first campus TCS in November 2011...i convey my heartly thanks to all the members of m4maths community who directly or indirectly helped me to get through TCS......special thanks to
admin for creating such a superb community
Manish Raj
2 years ago
this is important site for any one ,it changes my life...today i am part of tcs only because of M4ATHS.PUZZLE
Asif Neyaz
2 years ago
Thanku M4maths..due to u only, imade to TCS :D test on sep 15.
Harini Reddy
2 years ago
Big thanks to m4maths.com.
I cracked TCS..The solutions given were very helpful!!!
2 years ago
HI everyone ,
me and my friends vish,sube,shaf placed in TCS... its becoz of m4maths only .. thanks a lot..this is the wonderful website.. unless your help we might not have been able to place in TCS... and thanks
to all the users who clearly solved the problems.. im very greatful to you :)
2 years ago
Really thanks to m4maths I learned a lot... If you were not there I might not have been able to crack TCS.. love this site hope it's reputation grows exponentially...
2 years ago
Hello friends .I was selected in TCS. Thanx to M4Maths to crack apti. and my hearthly wishes that
the success rate of M4Math grow exponentially.
Again Thanx for all support given by M4Math during
my preparation for TCS.
and Best of LUCK for all students for their preparation.
2 years ago
thanks to M4MATHS..got selected in TCS..thanks for providing solutions to TCS puzzles :)
2 years ago
thousands of thnx to m4maths...
got selected in tcs for u only...
u were the only guide n i hv nvr done group study for TCS really feeling great...
thnx to all the users n team of m4maths...
3 cheers for m4maths
2 years ago
thousands of thnx to m4maths...
got selected in tcs for u only...
u were the only guide n i hv nvr done group study for TCS really feeling great...
thnx to all the users n team of m4maths...
3 cheers for m4maths
2 years ago
Thank U ...I'm placed in TCS.....
Continue this g8 work
2 years ago
thank you m4maths.com for providing a web portal like this.Because of you only i got placed in TCS,driven on 26/8/2011 in oncampus
raghu nandan
2 years ago
thanks a lot m4maths cracked TCS written n results are to be announced...is only coz of u... :)
V.V.Ravi Teja
3 years ago
thank u m4maths because of you and my co people who solved some complex problems for me...why because due to this only i got placed in tcs and hcl also........
Veer Bahadur Gupta
3 years ago
got placed in TCS ...
thanku m4maths...
Amulya Punjabi
3 years ago
Hi All,
Today my result for TCS apti was declared nd i cleared it successfully...It was only due to m4maths...not only me my all frnds are able to crack it only wid the help of m4maths.......it's just an
osum site as well as a sure shot guide to TCS apti......Pls let me know wt can be asked in the interview by MBA students.
Anusha Alva
3 years ago
a big thnks to this site...got placed in TCS!!!!!!
Oindrila Majumder
3 years ago
thanks a lot m4math.. placed in TCS
Pushpesh Kashyap
3 years ago
superb site, i cracked tcs
Saurabh Bamnia
3 years ago
Great site..........got Placed in TCS...........thanx a lot............do not mug up the sol'n try to understand.....its AWESOME.........
Gautam Kumar
3 years ago
it was really useful 4 me.................n finally i managed to get through TCS
Karthik Sr Sr
3 years ago
i like to thank m4maths, it was very useful and i got placed in tcs
rajesh dolai 3 Days ago plz send me ibm plaement paper english..
venkaiaha 3 Days ago cleared ibm 2 rounds thank you m4maths.com
Triveni 3 Days ago Thanks to m4maths.I got placed in IBM.Awsome work.Best of luck.
Lekshmi Narasimman MN 9 Days ago Thanks ton for this site . This site is my main reason for clearing cts written which happend on 5/4/2014 in chennai . Tommorrw i have my interview. Hope i will tel u
all a good news :) Thanks to almighty too :) !!
abhinay yadav 14 Days ago thank you M4maths for such awesome collection of questions. last month i got placed in techMahindra. i prepared for written from this site, many question were exactly same
as given here. bcz of practice i finished my written test 15 minutes before and got it. thanx allot for such noble work...
manasi 19 Days ago coz of this site i cud clear IBM's apti nd finally got placed in tcs thanx m4maths...u r a wonderful site :)
arnold 21 Days ago thank u m4maths and all its user for posting gud and sensible answers.
Nilesh singh 24 Days ago finally selected in TCS. thanks m4maths
MUDIT 26 Days ago Thank you team m4maths.Successfully placed in TCS.
Deepika Maurya 26 Days ago Thank you so much m4maths.. I cleared the written of IBM.. :) very good site.. thumps up !!
Rimi Das 1 Month ago Thanks to m4maths I got selected in Tech Mahindra.I was preparing for TCS 1st round since last month.Got interview call letter from there also...Really m4maths is the best site
for placement preparation...
Stephen raj 1 Month ago prepare from r.s.aggarwal verbal and non verbal reasoning and previous year questions from m4maths,indiabix and chetanas forum.u can crack it.
Stephen raj 1 Month ago Thanks to m4maths:) cracked infosys:)
Ranadip 1 Month ago i have been Selected in Tech Mahindra. All the quanti & reasoning questions are common from the placement papers of m4maths. So a big thanks to m4maths team & the people who
shares the placement papers.
Amit Das 1 Month ago I got selected for interview in TCS.Thank you very much m4maths.com.
PRAVEEN K H 1 Month ago I got placed in TCS :) Thanks a lot m4maths :)
Syed Ishtiaq 1 Month ago An Awesome site for TCS. Cleared the aptitude.
sara 1 Month ago I successfully cleared TCS aptitude test held on 8th march 2014.Thanks a lot m4maths.com plz guide for the technical round.
mounika devi mamidibathula 1 Month ago got placed in IBM.. this site is very useful, many questions repeated.. thanks alot to m4maths.com
Anisha Lakhmani 1 Month ago I got placed at infosys.......thanx to m4maths.com.......a awesum site......
Kusuma Saddala 2 Months ago Thanks to m4maths, i have place at IBM on feb 8th of this month
sangeetha 2 Months ago thanks to m4 maths because of this i clear csc written test
mahima srivastava 2 Months ago Placed at IBM. Thanks to m4maths. This site is really very helpful. 95% questions were from this site.
Surya Narayana K 2 Months ago I successfully cleared TCS aptitude test.Thanks a lot m4maths.com.
prashant gaurav 2 Months ago Got Placed In Infosys... Thanks of m4maths....
vishal 3 Months ago iam not placed in TCS...........bt still m4maths is a good site.
sameer 4 Months ago Thanx to m4 maths, because of that i able to crack aptitude test and now i am a part of TCS. This site is best for the preparation of placement papers.Thanks a
Sonali 4 Months ago THANKS a lot m4maths. Me and my 2 other roomies cleared the tcs aptitude with the help of this site.Some of the questions in apti are exactly same which i answered without even
reading the whole question completely.. gr8 work m4maths.. keep it up.
Kumar 5 Months ago m4maths is one of the main reason I cleared TCS aptitude. In TCS few questions will be repeated from previous year aptis and few questions will be repeated from the latest campus
drives that happened in various other colleges. So to crack TCS apti its enough to learn some basic concepts from famous apti books and follow all the TCS questions posted in m4maths. This is not
only for TCS but for all other companies too. According to me m4maths is best site for clearing apti. Kuddos to the creator of m4maths :)
YASWANT KUMAR CHAUDHARY 5 Months ago THANKS A LOT TO M4MATHS.due to m4maths today i am the part of TCS now.got offer letter now.
ANGELIN ALFRED 5 Months ago Hai friends, I got placed in L&T INFOTECH and i m visiting this website for the past 4 months.Solving placemetn puzzles from this website helped me a lot and
1000000000000s of thanks to this website.this website also encouraged me to solve puzzles.follw the updates to clear maths aps ,its very easy yar, surely v can crack it if v follow this website.
MALLIKARJUN ULCHALA 6 Months ago 2 days before i cleared written test just because of m4maths.com.thanks a lot for this community.
Madhuri 6 Months ago thanks for m4maths!!! bcz of which i cleared apti of infosys today.
DEVARAJU 6 Months ago Today my written test of TCS was completed.I answered many of the questions without reading entire question.Because i am one of the member in the m4maths. No words to praise
m4maths.so i simply said thanks a lot.
PRATHYUSHA BSN 7 Months ago I am very grateful to m4maths. It is a great site i have accidentally logged on when i was searching for an answer for a tricky maths puzzle. It heped me greatly and i am
very proud to say that I have cracked the written test of tech-mahindra with the help of this site. Thankyou sooo much to the admins of this site and also to all members who solve any tricky puzzle
very easily making people like us to be successful. Thanks a lotttt
Abhishek Ranjan 7 Months ago me & my rooom-mate have practiced alot frm dis site TO QUALIFY TCS written test.both of us got placed in TCS :) IT'S VERY VERY VERY HELPFUL N IMPORTANT SITE. do practice
n u'll surely succeed :)
Sandhya Pallapu 1 year ago Hai friends! this site is very helpful....i prepared for TCS campus placements from this site...and today I m proud to say that I m part of TCS family now.....dis site
helped me a lot in achieving this...thanks to M4MATHS!
vivek singh 2 years ago I cracked my first campus TCS in November 2011...i convey my heartly thanks to all the members of m4maths community who directly or indirectly helped me to get through
TCS......special thanks to admin for creating such a superb community
Manish Raj 2 years ago this is important site for any one ,it changes my life...today i am part of tcs only because of M4ATHS.PUZZLE
Asif Neyaz 2 years ago Thanku M4maths..due to u only, imade to TCS :D test on sep 15.
Harini Reddy 2 years ago Big thanks to m4maths.com. I cracked TCS..The solutions given were very helpful!!!
portia 2 years ago HI everyone , me and my friends vish,sube,shaf placed in TCS... its becoz of m4maths only .. thanks a lot..this is the wonderful website.. unless your help we might not have been
able to place in TCS... and thanks to all the users who clearly solved the problems.. im very greatful to you :)
vasanthi 2 years ago Really thanks to m4maths I learned a lot... If you were not there I might not have been able to crack TCS.. love this site hope it's reputation grows exponentially...
vijay 2 years ago Hello friends .I was selected in TCS. Thanx to M4Maths to crack apti. and my hearthly wishes that the success rate of M4Math grow exponentially. Again Thanx for all support given by
M4Math during my preparation for TCS. and Best of LUCK for all students for their preparation.
maheswari 2 years ago thanks to M4MATHS..got selected in TCS..thanks for providing solutions to TCS puzzles :)
GIRISH 2 years ago thousands of thnx to m4maths... got selected in tcs for u only... u were the only guide n i hv nvr done group study for TCS really feeling great... thnx to all the users n team of
m4maths... 3 cheers for m4maths
girish 2 years ago thousands of thnx to m4maths... got selected in tcs for u only... u were the only guide n i hv nvr done group study for TCS really feeling great... thnx to all the users n team of
m4maths... 3 cheers for m4maths
Aswath 2 years ago Thank U ...I'm placed in TCS..... Continue this g8 work
JYOTHI 2 years ago thank you m4maths.com for providing a web portal like this.Because of you only i got placed in TCS,driven on 26/8/2011 in oncampus
raghu nandan 2 years ago thanks a lot m4maths cracked TCS written n results are to be announced...is only coz of u... :)
V.V.Ravi Teja 3 years ago thank u m4maths because of you and my co people who solved some complex problems for me...why because due to this only i got placed in tcs and hcl also........
Veer Bahadur Gupta 3 years ago got placed in TCS ... thanku m4maths...
Amulya Punjabi 3 years ago Hi All, Today my result for TCS apti was declared nd i cleared it successfully...It was only due to m4maths...not only me my all frnds are able to crack it only wid the
help of m4maths.......it's just an osum site as well as a sure shot guide to TCS apti......Pls let me know wt can be asked in the interview by MBA students.
Anusha Alva 3 years ago a big thnks to this site...got placed in TCS!!!!!!
Oindrila Majumder 3 years ago thanks a lot m4math.. placed in TCS
Pushpesh Kashyap 3 years ago superb site, i cracked tcs
Saurabh Bamnia 3 years ago Great site..........got Placed in TCS...........thanx a lot............do not mug up the sol'n try to understand.....its AWESOME.........
Gautam Kumar 3 years ago it was really useful 4 me.................n finally i managed to get through TCS
Karthik Sr Sr 3 years ago i like to thank m4maths, it was very useful and i got placed in tcs
Latest User posts (More)
Maths Quotes (More)
"It is impossible to be a mathematician without being a poet in soul." Sofia Kovalevskaya
"MATHEMATICS is a great motivator for all humans.. Because its career starts with "ZERO" but it never end(INFINITY).. " Kumar Purnendu
"But mathematics is the sister, as well as the servant, of the arts and is touched with the same madness and genius." Harold Marston Morse
""IF YOU DON'T KNOW MATHS , IT MEAN YOU ARE FAR AWAY FROM LIFE."" Ayush
"A man whose mind has gone astray should study mathematics." Francis Bacon
"every maths problem has 108 types solution." Rakshith Shetty...
""5/3rds of all people just don't get fractions"" Unknown
Latest Placement Puzzle (More)
"What is the next number of the following sequence
16,22,26,38,62,74,...... "
UnsolvedAsked In: Self
"the ratios of the monthly income of A and B are in the ratio of 3:4.The ratio of the monthly
expenditures of A and B is 4:5 .Find the ratio of their savings
UnsolvedAsked In: CAT
"x,yand z are three quantities.X varies inversely with Y when Z is constant.Y varies inversely with Z when X is constant.When Y=8,and Z=7,X=30.Find X if Y=16 and Z=21."
UnsolvedAsked In: AMCAT
"It is impossible to be a mathematician without being a poet in soul." Sofia Kovalevskaya
"MATHEMATICS is a great motivator for all humans.. Because its career starts with "ZERO" but it never end(INFINITY).. " Kumar Purnendu
"But mathematics is the sister, as well as the servant, of the arts and is touched with the same madness and genius." Harold Marston Morse
""IF YOU DON'T KNOW MATHS , IT MEAN YOU ARE FAR AWAY FROM LIFE."" Ayush
"A man whose mind has gone astray should study mathematics." Francis Bacon
"What is the next number of the following sequence 16,22,26,38,62,74,...... " UnsolvedAsked In: Self
"the ratios of the monthly income of A and B are in the ratio of 3:4.The ratio of the monthly expenditures of A and B is 4:5 .Find the ratio of their savings " UnsolvedAsked In: CAT
"x,yand z are three quantities.X varies inversely with Y when Z is constant.Y varies inversely with Z when X is constant.When Y=8,and Z=7,X=30.Find X if Y=16 and Z=21." UnsolvedAsked In: AMCAT
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Mining For The 'Prime' Jewels Of Numbers
In Depth
Web Resources
Set To Type...
NPR asked the typographers at Hoefler & Frere-Jones to estimate how long this Mersenne prime would be if typed out — all 12,987,189 digits. The answer, it turns out, depends not only on the font but
also the kerning, or the space between characters.
Set to 12 point Mercury Text: 17 miles, 202 feet, 1.5 inches. And to 12 point Gotham Book? That'd be 20 miles, 3,131 feet, 1.75 inches.
When they're not being used to send e-mails or play solitaire, about 50,000 personal computers around the world are engaged in a search — for the world's largest prime number.
Prime numbers — like 2, 3, 5 and 7 — are numbers that are divisible only by the number 1 and by themselves.
The largest prime found to date is nearly 13 million digits long. To get an idea of just how big this prime number is, if you write 10 digits per inch — all 12,978,189 of them — the number would
extend for 20.45 miles. The number is currently the world's largest prime. But there's always a larger one to find.
These gargantuan primes fall under a special category known as Mersenne primes, named for a 17th century French theologian who made some predictions about them. (His early predictions ultimately
turned out to be wrong.)
Computational Torture
The current reigning champ of Mersenne primes was discovered last summer as part of a program called the Great Internet Mersenne Prime Search, or GIMPS.
Apparently, Mersenne primes are easier to find than other primes, and George Woltman, a software designer in Florida and the man behind GIMPS, has written a free, downloadable program to search for
them. But you need a lot of computing power.
"It takes about two or three weeks to test a single number," he says. "And everybody is plugging away, trying to find yet another prime number."
Chris Caldwell, a mathematician at the University of Tennessee, Martin, says that the main obstacle in proving that these numbers are prime is just doing the arithmetic with numbers that are millions
of digits long. Caldwell says there is a formula for testing whether a large number is a Mersenne prime, but it's computationally intense.
"Not only do you have to multiply a 13 million-digit number by a 13 million-digit number, but you have to do that about 13 million times," Caldwell says. "And that just takes a tremendous amount of
GIMPS has been plugging away at this for 13 years now, and has found 12 Mersennes so far.
'The Jewels Of Number Theory'
So why are people anxious to find another, larger Mersenne prime? The response from many in the math community is generally the same: because.
"Mersennes, in a way, are kind of like a large diamond," Caldwell explains. When he went to Washington, he says, he took his kids to see the Hope Diamond. That's the 45-carat diamond that sits in a
special case in the National Museum of Natural History, usually with crowds around it.
"Nobody there looking at the Hope Diamond ever asks, 'Why did they bother to dig it up?' or 'What is it good for?' — even though it really isn't good for much other than to just hang there and people
to look at," Caldwell says. "And in many ways the Mersennes play that same role — that they really are the jewels of number theory."
See The Largest Known Prime, All 13 Million Digits
We all think we remember prime numbers from grade school, but just in case your memory is a little hazy, here's a quick refresher.
A prime is a number that can only be divided by two other whole numbers: itself and 1.
So, 1? Not a prime — it can only be divided by 1. You need two divisors to qualify for prime status.
Marked in orange are some of the easy primes — because they don't take much brain power to figure out:
But then, there are the special primes. These numbers come from a variety of simple equations, but present mind-bogglingly large numbers. Finding primes was sort of an elite hobby for some of the
early math greats, like Pierre de Fermat, Carl Gauss and Sophie Germain.
But the most celebrated of all primes is the Mersenne — what mathematician Chris Caldwell calls "the jewels of number theory."
Rare, Huge Numbers
Mersenne primes, named after 17th-century mathematician Marin Mersenne, are numbers derived from the simple equation:
The number n is a prime, and the result is prime. And what makes the Mersenne primes so interesting is how rare they are. And their gargantuan size.
Currently, a Mersenne prime clocking in at nearly 13 million digits holds the record of being the largest prime number ever found. You can take a look at just a snippet of the number here. But if you
want to see all 13 million digits, you can download this whopping 17-megabyte text file. (That's almost 3,500 single-spaced pages in a Word document, at 12-point Times New Roman font.)
And here's a glimpse at how quickly Mersennes ramp up in size:
Notice that as the value of the number n — the exponents 2, 3, 5, 7, 13, 17 and 19 in the example above — increases, the value of the Mersenne prime skyrockets. If we plot the value of the Mersenne
prime vs. the value of the exponent, here's what the pattern looks like:
And here's an easier way to think of that Mersenne prime at the top right of the chart: It's more than a trillion trillion trillion. You get that number with this equation:
So just using an exponent of 127 gives you that huge number. But the largest Mersenne prime's exponent is 43,112,609.
To find such a large number is beyond the brain power of anyone. Microsoft Excel can't even calculate any Mersenne prime with an exponent larger than 607. If you try, it gives you a nice #NUM! error
— that's Excelese for FAIL.
You need massive computers to do the number crunching. So a team called the Great Internet Mersenne Prime Search, or GIMPS, is farming the computing power out to the Internet. Of the 46 Mersenne
primes found so far, GIMPS has discovered 12.
Users can lend their desktop computing power out to a program using a type of software called distributed computing. This type of large-scale marshaling of computing power is not new to the Internet.
Other groups, like the Search For Extraterrestrial Intelligence, or SETI@home, use personal computers for data crunching, as does the Folding@home program, based at Stanford University, whose goal is
to learn more about how the proteins found in living organisms fold into complex shapes and interact with other biological molecules.
So why bother searching for these primes? Any math geek will tell you that it's just cool and they're in it for the chase. But the effort has been reciprocal, too. As they search for these
ever-larger primes, they have also been able to contribute important theorems and principles to the field of mathematics.
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Why is the integral of the second chern class an integer?
up vote 9 down vote favorite
I'm currently reading the paper "Holonomy, the Quantum Adiabatic Theorem, and Berry's Phase" by Barry Simon.
Imagine a vector bundle with a connection $\nabla$. For simplicity, we assume that this is a $U(1)$ vector bundle. Parallel transport gives rise to the holonomy group, which assigns to each curve $C$
a number $e^{i\gamma(C)}$ that indicates how a vector is "rotated" when transporting it along the curve. In turns out that the phase change $\gamma(C)$ can be expressed as an integral of the
curvature form over any surface $S$ that delimits the curve, $C = \partial S$,
$$ \gamma(C) = \int_{S} F^{\nabla} .$$
I am interested in the integral of the curvature form over the whole manifold, which turns out to be an integer multiple of $2\pi$,
$$ \int_{M} F^{\nabla} = 2\pi k, k\in\mathbb{Z}$$
Simon notes that this "standard fact" is a consistency condition on the holonomy group. I can understand that: integrating over the whole manifold is like taking the holonomy of the constant path,
which must be the identity.
What I would like to understand is the generalization to higher Chern classes. For instance,
Why is the integral of the second Chern form an integer multiple of $4\pi^2$?
$$ \int_{M} F^{\nabla}\wedge F^{\nabla} = 4\pi^2 k, k\in\mathbb{Z}$$
I have a pedestrian proof for special cases, but I would like to understand a general reason behind this phenomenon. Is there a "higher holonomy" at work here?
Obviously, my knowledge of vector bundles and characteristic classes is rather limited. I can find my way around the book "From Calculus to Cohomology", but have by no means absorbed all the
material. Basically, my question is why the Chern classes defined via connections are normalized with a factor of $1/(2\pi)^k$.
chern-classes vector-bundles connections
2 A standard trick in the trade is the splitting principle, which says in effect that you can pretend that your vector bundle (with connection) is a sum of line bundles. There ways to justify it.
Apply it, then you'll see that the factors of $2\pi (i)$ in the higher classes will have to multiply. – Donu Arapura Mar 28 '11 at 16:58
The question is already restricted to complex line bundles. Or are you saying that there is a way to interpret the second form $F^{\nabla}\wedge F^{\nabla}$ as a form of lower degree in a
higher-dimensional bundle? – Greg Graviton Mar 28 '11 at 17:25
1 Greg, you can't mean this! $c_2$ and higher for a line bundle are zero in cohomology. In fact, your integrand is zero. You want to have at least rank $2$, and take $\int p(F\wedge F)$ where $p$ is
an elementary symmetric function. – Donu Arapura Mar 28 '11 at 17:37
2 The title of the question misrepresents the content. Its not about the second Chern class. In fact the question seems to be about the Chern-Weil formula for characteristic classes of a complex
vector bundle. Here is a more succinct statement. Let $t$ be a formal variable, and consider $$det(\frac{it\Omega}{2\pi}+Id)$$ where $\Omega$ is the curvature form of a connection on a complex
vector bundle over a smooth manifold. Why do the coefficients of $t$ lie in the image of the integral cohomology of the manifold inside the DeRham cohomology? – Charlie Frohman Mar 29 '11 at 12:48
1 Donu's remark at the beginning answers the question modulo checking that the integral over $i\Omega/2\pi$ over $\mathbb{C}P(1)$ for the standard connection on the canonical line bundle over $\
mathbb{C}P(n)$ is $ 1$. – Charlie Frohman Mar 29 '11 at 12:56
show 3 more comments
3 Answers
active oldest votes
I realize that I may as well write an answer rather than a series of comments. Although Jessica has given a good answer, I'll try to say this as concretely as possible, since I now think I
understand the question more clearly. The question was actually about the integrality of $$\frac{1}{4\pi^2}\int_M F\wedge F$$ where $F$ is the curvature of line bundle $V$ on a $4$-manifold
$M$. This is what mathematicians (I'm assuming you're a physicist) would call $c_1(V)^2$. The first thing is observe that $c_1(V)\in H^2(M,\mathbb{Z})$, and that it's image in de Rham
cohomology is given by $1/(2\pi i)[F]$. To see this in explicit terms, note that the classifying space for line bundles in $\mathbb{C}\mathbb{P}^\infty$. This implies that the $V$ is the
up vote 8 pull back of the tautological bundle under a $C^\infty$ map $f:M\to \mathbb{C}\mathbb{P}^N$, for $N\gg 0$. Working on projective space, we can check integrality of the class $1/(2\pi i)[F]$
down vote by doing a direct calculation to see that this integrates to $1$ over a complex line (aka $2$-sphere). This suffices because the line generates $H_2(\mathbb{C}\mathbb{P}^N)$. After this,
accepted $c_1(V)^2=-1/4\pi^2[F]^2$ is automatically integral. That's it.
Postscript: If you are unhappy with the last part, you can replace $f$ with its composition with a generic projection to obtain $f:M\to \mathbb{C}\mathbb{P}^2$. Then your integral becomes
the degree of $f$ which is certainly an integer. Hopefully, you can take it from here.
D'oh! I mixed up the nomenclature. $F \wedge F$ would be the second Chern character class according to "From Calculus to Cohomology". I'm not familiar with classifying spaces (why is
there a map $f : M \to \mathbb{CP}^N$ for any manifold $M$), but I don't care much about that. – Greg Graviton Mar 29 '11 at 16:44
What seems more interesting is the integrality property. Assuming I only know de-Rham cohomology, could you elaborate on how I can detect whether a differential form from $H^k(M,\mathbb
{C})$ is already a member of $H^k(M,\mathbb{Z})$? Is there a short reason why the wedge (cup) product of two integral forms is again integral? – Greg Graviton Mar 29 '11 at 16:44
3 If you only know deRham cohomology, then you don't know what $H^k(X, \mathbb{Z})$ means! OK, that's a bit unfair. It looks like in practice, you are interested in the subspace of $\
omega$ in $H^k(X, \mathbb{R})$ such that, for every $k$-cycle $\sigma$, we have $\int_{\sigma} \omega \in \mathbb{Z}$. This is the image of $H^k(X, \mathbb{Z})$ in $H^k(X, \mathbb{R})$.
So it sounds like what you want to know is why wedge product preserves this property of differential forms. – David Speyer Mar 29 '11 at 19:33
I'd be curious to see a direct proof of this, which doesn't go through the construction of $H^k(X, \mathbb{Z})$. Why don't you ask this as a separate question over on
math.stackexchange.com ? (It will get closed here, because people here will have no problem assuming that integral cohomology exists.) – David Speyer Mar 29 '11 at 19:34
Thank you, David, I have done as you suggested. – Greg Graviton Mar 29 '11 at 20:17
add comment
Let $V$ be a complex vector bundle on a manifold $M$. Chern classes can be defined by topological means (see Milnor's book on characteristic classes), which yields elements $c_k(V) \in H^
{2k}(M;\mathbb{Z})$. The normalization in the Chern-Weil theory is chosen so that the associated elements of de Rham cohomology groups $H^{2k}(M;\mathbb{R})$ agree with the integral
up vote 13 elements, and thus integrate to give integers.
down vote
4 That is certainly true, but kind of dodges the question. One would have to show why the two agree up to a constant factor and then one would have to determine the constant factor to be
precisely $(2\pi)^k$, and not, say $3\pi^{k/2}$. Both are currently beyond my understanding. – Greg Graviton Mar 28 '11 at 17:08
7 Let's say that you knew everything (functoriality, Whitney sum...) but the correct normalization. Then you'd have to calculate essentially one example: the universal bundle on a
Grassmanian to get the correct constant. To do carry this out, pull it up to the flag manifold (you won't lose anything, since cohomology injects), split it has a sum of line bundles,
and apply Whitney sum. That's what I meant in my previous comment. – Donu Arapura Mar 28 '11 at 17:30
1 Or, as the poser of the question is familiar with "From Calculus to Cohomology" you can quote Theroem 18.4 on page 184. – Charlie Frohman Mar 29 '11 at 15:44
add comment
Take a look at Appendix C in Milnor's book on characteristic classes. Essentially what is going on is that if you have a complex line bundle $L$ with connection $\nabla$ and curvature form
$K_\nabla$, then the cohomology class of $\sigma_r(K_\nabla)$ is equal to $(2\pi i)^r c_r(L)$. Here $\sigma_r$ is the $r$th elementary symmetric function on the eigenvalues of the (matrix of
the) connection.
up vote The equality $\sigma_1(K_\nabla) = 2\pi i c_1(L)$ is rather transparent in case $L$ is a line bundle over a surface $S$ (as in the OP). Indeed, $\sigma_1 = \text{trace}$, and so what's being
5 down said is that $K_\nabla = 2\pi i c_1(L)$. And why is this true? Well, $K_\nabla$ is a closed $2$-form on $S$ that represents a characteristic cohomology class in $H^2(S;\mathbb{C})$, and
vote therefore must be some multiple $a c_1(L)$ of the first Chern class. This constant $a$ is independent of $L$. So to compute it, all you need to do is work out some specific example. The
formula $$ \int_S F^\nabla = 2\pi i k $$ given in the OP (i.e. the Gauss--Bonnet formula!) does just that. It follows that $a=2\pi i$.
1 Thank you for pointing me to appendix C in Milnor's book! – Greg Graviton Mar 29 '11 at 20:18
add comment
Not the answer you're looking for? Browse other questions tagged chern-classes vector-bundles connections or ask your own question.
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Variation ranks of communication matrices and lower bounds for depth two circuits having symmetric gates with unbounded fan--in
Results 1 - 10 of 34
- Computational Complexity , 1992
"... . In this paper we study small depth circuits that contain threshold gates (with or without weights) and parity gates. All circuits we consider are of polynomial size. We prove several results
which complete the work on characterizing possible inclusions between many classes defined by small depth c ..."
Cited by 89 (7 self)
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. In this paper we study small depth circuits that contain threshold gates (with or without weights) and parity gates. All circuits we consider are of polynomial size. We prove several results which
complete the work on characterizing possible inclusions between many classes defined by small depth circuits. These results are the following: 1. A single threshold gate with weights cannot in
general be replaced by a polynomial fan-in unweighted threshold gate of parity gates. 2. On the other hand it can be replaced by a depth 2 unweighted threshold circuit of polynomial size. An
extension of this construction is used to prove that whatever can be computed by a depth d polynomial size threshold circuit with weights can be computed by a depth d + 1 polynomial size unweighted
threshold circuit, where d is an arbitrary fixed integer. 3. A polynomial fan-in threshold gate (with weights) of parity gates cannot in general be replaced by a depth 2 unweighted threshold circuit
of polynomial size...
, 2000
"... We investigate the computational power of depth two circuits consisting of MOD r --gates at the bottom and a threshold gate with arbitrary weights at the top (for short, threshold--MOD r
circuits) and circuits with two levels of MOD gates (MOD p -MOD q circuits). In particular, we will show ..."
Cited by 56 (4 self)
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We investigate the computational power of depth two circuits consisting of MOD r --gates at the bottom and a threshold gate with arbitrary weights at the top (for short, threshold--MOD r circuits)
and circuits with two levels of MOD gates (MOD p -MOD q circuits). In particular, we will show the following results. (i) For all prime numbers p and integers q; r, it holds that if p divides r but
not q then all threshold--MOD q circuits for MOD r have exponentially many nodes. (ii) For all integers r, all problems computable by depth two fAND;OR;NOTg-- circuits of polynomial size have
threshold--MOD r circuits with polynomially many edges. (iii) There is a problem computable by depth 3 fAND;OR;NOTg--circuits of linear size and constant bottom fan--in which for all r needs
threshold--MOD r circuits with exponentially many nodes. (iv) For p; r different primes, and q 2; k positive integers, where r does not divide q; every MOD p k -MOD q circuit for MOD r has e...
- Computational Complexity , 1994
"... . Define the MODm -degree of a boolean function F to be the smallest degree of any polynomial P , over the ring of integers modulo m, such that for all 0-1 assignments ~x, F (~x) = 0 iff P (~x)
= 0. We obtain the unexpected result that the MODm -degree of the OR of N variables is O( r p N ), wher ..."
Cited by 56 (6 self)
Add to MetaCart
. Define the MODm -degree of a boolean function F to be the smallest degree of any polynomial P , over the ring of integers modulo m, such that for all 0-1 assignments ~x, F (~x) = 0 iff P (~x) = 0.
We obtain the unexpected result that the MODm -degree of the OR of N variables is O( r p N ), where r is the number of distinct prime factors of m. This is optimal in the case of representation by
symmetric polynomials. The MOD n function is 0 if the number of input ones is a multiple of n and is one otherwise. We show that the MODm -degree of both the MOD n and :MOD n functions is N\Omega\
Gamma1/ exactly when there is a prime dividing n but not m. The MODm -degree of the MODm function is 1; we show that the MODm -degree of :MODm is N\Omega\Gamma30 if m is not a power of a prime, O(1)
otherwise. A corollary is that there exists an oracle relative to which the MODmP classes (such as \PhiP) have this structure: MODmP is closed under complementation and union iff m is a prime power,
- In Proc. 36th , 1996
"... The rigidity of a matrix measures the number of entries that must be changed in order to reduce its rank below a certain value. The known lower bounds on the rigidity of explicit matrices are
very weak. It is known that stronger lower bounds would have implications to complexity theory. We consider ..."
Cited by 48 (3 self)
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The rigidity of a matrix measures the number of entries that must be changed in order to reduce its rank below a certain value. The known lower bounds on the rigidity of explicit matrices are very
weak. It is known that stronger lower bounds would have implications to complexity theory. We consider restricted variants of the rigidity problem over the complex numbers. Using spectral methods, we
derive lower bounds on these variants. Two applications of such restricted variants are given. First, we show that our lower bound on a variant of rigidity implies lower bounds on size-depth
tradeoffs for arithmetic circuits with bounded coefficients computing linear transformations. These bounds generalize a result of Nisan and Wigderson. The second application is conditional; we show
that it would suffice to prove lower bounds on certain restricted forms of rigidity to conclude several separation results such as separating the analogs of PH and PSPACE in communication complexity
theory. Our res...
, 1997
"... The 1980's saw rapid and exciting development of techniques for proving lower bounds in circuit complexity. This pace has slowed recently, and there has even been work indicating that quite
different proof techniques must be employed to advance beyond the current frontier of circuit lower bounds. Al ..."
Cited by 30 (3 self)
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The 1980's saw rapid and exciting development of techniques for proving lower bounds in circuit complexity. This pace has slowed recently, and there has even been work indicating that quite different
proof techniques must be employed to advance beyond the current frontier of circuit lower bounds. Although this has engendered pessimism in some quarters, there have in fact been many positive
developments in the past few years showing that significant progress is possible on many fronts. This paper is a (necessarily incomplete) survey of the state of circuit complexity as we await the
dawn of the new millennium.
, 2001
"... Abstract We use a Ramsey-theoretic argument to obtain the firstlower bounds for approximations over Zm by nonlinearpolynomials: ffl A degree-2 polynomial over Zm (m odd) mustdiffer from the
parity function on at least a ..."
Cited by 29 (0 self)
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Abstract We use a Ramsey-theoretic argument to obtain the firstlower bounds for approximations over Zm by nonlinearpolynomials: ffl A degree-2 polynomial over Zm (m odd) mustdiffer from the parity
function on at least a
- PROCEEDINGS OF THE 3RD SWAT SCANDINAVIAN WORKSHOP ON ALGORITHM THEORY, HELSINKI, FINLAND (LNCS 621 , 1992
"... In this talk we will consider various classes defined by small depth polynomial size circuits which contain threshold gates and parity gates. We will describe various inclusions between many
classes defined in this way and also classes whose definitions rely upon spectral properties of Boolean fu ..."
Cited by 20 (2 self)
Add to MetaCart
In this talk we will consider various classes defined by small depth polynomial size circuits which contain threshold gates and parity gates. We will describe various inclusions between many classes
defined in this way and also classes whose definitions rely upon spectral properties of Boolean functions.
, 1998
"... We show that at least =r) entries must be changed in an arbitrary (generalized) Hadamard matrix in order to reduce its rank below r. This improves upon the previously known bound ) [6]. If we
additionally know that these changes are bounded from above in their absolute values by so ..."
Cited by 19 (0 self)
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We show that at least =r) entries must be changed in an arbitrary (generalized) Hadamard matrix in order to reduce its rank below r. This improves upon the previously known bound ) [6]. If we
additionally know that these changes are bounded from above in their absolute values by some n=r, we prove another bound on their number. This improves upon the bound ) from [13].
, 2000
"... . A set F of Boolean functions is called a pseudorandom function generator (PRFG) if communicating with a randomly chosen secret function from F cannot be efficiently distinguished from
communicating with a truly random function. We ask for the minimal hardware complexity of a PRFG. This question is ..."
Cited by 11 (1 self)
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. A set F of Boolean functions is called a pseudorandom function generator (PRFG) if communicating with a randomly chosen secret function from F cannot be efficiently distinguished from communicating
with a truly random function. We ask for the minimal hardware complexity of a PRFG. This question is motivated by design aspects of secure secret key cryptosystems. Such cryptosystems should be
efficient in hardware, but often are required to behave like PRFGs. By constructing efficient distinguishing schemes we show for a wide range of basic nonuniform complexity classes, induced by depth
restricted branching programs and several types of constant depth circuits (including TC 0 2 ), that they do not contain PRFGs. On the other hand we show that the PRFG proposed by Naor and Reingold
in [24] consists of TC 0 4 -functions. The question if TC 0 3 -functions can form PRFGs remains as an interesting open problem. We further discuss relations of our results to previous work on
cryptographic ...
- In Proc. 26th Ann. ACM Symp. Theor. Comput , 1994
"... : We prove that any depth--3 circuit with MOD m gates of unbounded fan-in on the lowest level, AND gates on the second, and a weighted threshold gate on the top needs either exponential size or
exponential weights to compute the inner product of two vectors of length n over GF(2). More exactly we p ..."
Cited by 10 (1 self)
Add to MetaCart
: We prove that any depth--3 circuit with MOD m gates of unbounded fan-in on the lowest level, AND gates on the second, and a weighted threshold gate on the top needs either exponential size or
exponential weights to compute the inner product of two vectors of length n over GF(2). More exactly we prove that log(wM ) = \Omega\Gamma n), where w is the sum of the absolute values of the
weights, and M is the maximum fan--in of the AND gates on level 2. Setting all weights to 1, we have got a trade--off between the numbers of the MOD m gates and the AND gates. By our knowledge, this
is the first trade--off result involving hard--to--handle MOD m gates. In contrast, with n AND gates at the bottom and a single MOD 2 gate at the top one can compute the inner product function. The
lower--bound proof does not use any monotonicity or uniformity assumptions, and all of our gates have unbounded fan--in. The key step in the proof is a random evaluation protocol of a circuit with
MOD m gates. ...
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2nd order differential equation - variation of parameters
March 30th 2009, 11:50 PM #1
Mar 2009
2nd order differential equation - variation of parameters
I would like to know how to solve the following:
Solve the differential equation using the method of varaition of parameters.
y" - 3y' + 2y = 1/(1+e^-x)
What I have done till now (not really able to continue):
- solution to y" - 3y' + 2y = 0 is c1e^x+c2e^2x
-y1 = e^x & y2 = e^2x
-yp = u1e^x + u2e^2x
However, I am not sure if I have u2 correctly:
u1 = ln (1+e^-x)
u2 = -(1+e^-x) + ln (1+e^-x) where I have integral of (e^-2x)/(1+e^-x)
can someone please check with me and help me solve the following please.
Thanks in advance!
I would like to know how to solve the following:
Solve the differential equation using the method of varaition of parameters.
y" - 3y' + 2y = 1/(1+e^-x)
What I have done till now (not really able to continue):
- solution to y" - 3y' + 2y = 0 is c1e^x+c2e^2x
-y1 = e^x & y2 = e^2x
-yp = u1e^x + u2e^2x
However, I am not sure if I have u2 correctly:
u1 = ln (1+e^-x)
u2 = -(1+e^-x) + ln (1+e^-x) where I have integral of (e^-2x)/(1+e^-x)
can someone please check with me and help me solve the following please.
Thanks in advance!
When I did it, I got your u1 but for u2
$u_2 = -e^{-x} + \ln | 1 + e^{-x}|$
but I don't think it matters in the end b/c when you expand
$u_1 e^x + u_2 e^{2x}$
the first term in u2 (either yours or mine) can be absorbed into the complementary solution.
March 31st 2009, 06:25 AM #2
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Program Correctness
Robert W Floyd (Source: Stanford Computer Science [4])
We present a project for students on proving the correctness of a program, studied by reading excerpts from the pioneering paper of Robert W. Floyd (1936-2001) on “Assigning meanings to programs” [1
Another pioneering paper in program correctness was C. A. R. Hoare’s “An axiomatic basis for computer programming” [2]. Almost all treatments of program correctness in advanced textbooks are based on
Hoare’s logical or axiomatic approach, in which mathematical axioms and inference rules are employed. But as Hoare admitted, “The formal treatment of program execution presented in this paper is
clearly derived from Floyd” [2, p. 583] and the treatment “is essentially due to Floyd but is applied to texts rather than flowcharts” [2, p. 577]. That is, conceptually Hoare’s method is no
different than Floyd’s. However, Hoare’s axiomatic approach, which is presented with the notation and terminology of axioms and inference rules, may be quite difficult for an upper level computer
science undergraduate student. In practice, Hoare’s technique, while based on axioms and inference rules, is often carried out by annotating a program with assertions.
In Floyd’s approach, programs are presented as flowcharts. One may wonder if flowchart programs are realistic, and whether Floyd’s technique applies to programs that people write nowadays. In this
project, we propose to replace flowchart programs by assembly-like programs with the use of “goto” statements. Each statement is accompanied by an assertion.
By adopting assembly-like programs as our preferred programming style, we are able to present the correctness proof technique in a simple and unified manner without the need to revise the technique
for each new programming language. We assume that a junior level computer science student is able to convert mechanically a program in a high level language into an assembly-like program. As the
mechanical conversion preserves the same program execution behavior, we can establish the correctness of the original program in a high level language by proving the correctness of the derived
assembly-like program.
Our project, Program Correctness, is ready for students, and the Latex source is also available for instructors who may wish to modify the project for students. The comprehensive “Notes to the
Instructor” presented next are also appended to the project itself. Our project is part of a larger collection published in Convergence. For additional projects, see Primary Historical Sources in the
Classroom: Discrete Mathematics and Computer Science.
Notes to the Instructor
The project is designed to focus on the key ideas in program correctness, rather than on the formal aspects of the theory. Terminology like axioms, inferences, and deductive systems is deliberately
avoided. Emphasis is placed on the practice of proving the partial correctness of interesting programs. It is the author’s belief that the students will appreciate better the power of program
correctness through the analysis of seemingly simple and yet non-trivial programs. Furthermore, the students will hopefully learn how to write better programs as a result of this study.
The project is suitable for an undergraduate upper-level Algorithms course or Programming Languages course. Most of the work can be completed by the students individually.
Floyd’s flowchart program for correctness proof is an excellent example. The instructor should go over this example slowly. Students should be given ample time to analyze the annotations given in the
flowchart program. The objective is to develop an informal understanding through the induction principle that the annotations together constitute a partial correctness proof.
The correctness proof (Task 13) of the Partition procedure is very challenging. It may be skipped if time is limited, or if the students are finding it too difficult. Students can also tackle Task 13
collaboratively. Work can be assigned in three parts (Tasks 1-10, Tasks 11-12, and Task 13) with one week given for each part. Task 14 can be considered as extra credit. After students’ answers for
the first part (Tasks 1-10) have been turned in, the instructor can discuss the uses of back substitution (as given in the verification condition for the assignment operator) that are required for
the answers for Tasks 9 and 10, so as to be sure that the students are ready to take on the tasks given in Sections 5, 6 and 7.
The project does not cover termination proof. Unlike partial correctness proof, the concept of termination proof is quite straightforward. Students will find good discussions of termination proof in
their textbooks.
Finally, students may be interested to know that both Robert Floyd and Tony Hoare were awarded the Alan M. Turing Award, generally considered to be the highest award in computer science, Floyd in
1978 [5] and Hoare in 1980 [3].
Download the project Program Correctness as a pdf file ready for classroom use.
Download the modifiable Latex source file for this project.
For more projects, see Primary Historical Sources in the Classroom: Discrete Mathematics and Computer Science.
[1] Floyd, Robert W., “Assigning meanings to programs,” Proceedings of Symposia in Applied Mathematics, Vol. 19 (1967), pages 19–32.
[2] Hoare, C. A. R., “An axiomatic basis for computer programming,” Communications of the ACM, Vol. 12, No. 10 (1969), pages 576–583.
[3] Jones, Cliff, “C. Antony (“Tony”) R. Hoare, United Kingdom – 1980,” Association for Computing Machinery (ACM) A. M. Turing Award website, 2012,
http://amturing.acm.org/award_winners/
[4] Levy, Dawn, “Professor Robert W. Floyd,” In Memorium, Stanford University Computer Science website, September 2001,
http://www-cs.stanford.edu/faculty/memoriam
http://www-cs.stanford.edu/content
[5] “Robert (Bob) W Floyd, United States – 1978,” Association for Computing Machinery (ACM) A. M. Turing Award website, 2012,
http://amturing.acm.org/award_winners/hoare_4622167.cfm
The development of curricular materials for discrete mathematics has been partially supported by the National Science Foundation's Course, Curriculum and Laboratory Improvement Program under grants
DUE-0717752 and DUE-0715392 for which the authors are most appreciative. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not
necessarily reflect the views of the National Science Foundation.
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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.
This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.
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Total Moment Forces on a Drum
February 3rd 2010, 04:41 PM #1
Jan 2010
Total Moment Forces on a Drum
See figure attached.
I tried to split all the forces and calculate the perpendicular distances from the point E. I did this for the 2 forces in the 2nd and 4th quadrant but I still don't have enough information to
determine alpha.
Any ideas?
EDIT: Update, I think I managed to solve it by simply expressing the perpendicular distances of the 130N force in terms of alpha.
Anyone care to check my answer?
I got a value of 39.1 degrees for alpha.
Thanks again.
Last edited by jegues; February 3rd 2010 at 05:12 PM.
Bump, still looking for verification.
February 7th 2010, 07:53 PM #2
Jan 2010
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Physics at the Gym
Physics at the Gym By Adam Weiner Posted 05.20.2009
Who isn't amused by the rare and impressive science-savvy party trick? One that involves the potential to risk death death by flinging yourself Superman-like at a bouncy training ball, only to have
it pop you back up in an amazingly graceful backflip? Before you cry "Sir Isaac Newton!," here are the physics behind this seemingly impossible stunt.
Exhibit A: The trick at hand, looped for your enjoyment:
This trick makes use of the principle of conservation of momentum, in both linear and angular form! Now you can oooh and aaah. According to this principle, in the absence of external forces on an
object or a system of objects the total momentum of the system will remain constant.
In order to get the ball, and “the guy”, moving in the first place we need external forces. But once they’re in motion, the combined momentum of the guy and ball just before and just after the
collision will be the same. Now, neither “the guy’s”, nor the ball’s, momentum changes as a result of the collision, but as a system of two objects the total is constant. However much momentum one
object (like “the guy”) gains is exactly equal to the amount the other (the ball) loses.
Let’s see how this plays out in this particular stunt. Before the collision, the total momentum of the system consists of the combined linear momentum of the ball and the guy, as well as the ball’s
angular momentum (due to the fact that it’s rolling). Linear momentum (p) of an object is equal to the mass of the object multiplied by the velocity of its center of mass. Mathematically
p = mv
and as we mentioned above, the total momentum is conserved in the collision. So:
mguy vguy + mball vball (before collision) = mguy vguy + mball vball (after collision).
Notice how the guy’s forward motion essentially stops as a result of the collision. (We’ll get to the rotation in a second.) His linear momentum decreases. In order to conserve momentum, the velocity
of the ball must increase as a result of the collision, which is exactly what happens. The guy experiences a force from the ball reducing his momentum, but Newton’s third law-- the law of action and
reaction-- says that the guy pushes with an equal force back on the ball. The ball therefore increases its momentum by exactly the same amount as the guy’s decreases. Total momentum is constant.
Physics in action!
Angular momentum (L) has to do with the momentum of a rotating object.
L = Iω
I is something called the rotational inertia of the object and depends not only on its mass but where the mass is distributed, and ω is the angular velocity (or rotation rate). To change the angular
momentum of an object you need a torque, which, for our purposes, we can think of as a force acting away from the center of an object’s mass. Analogous to the linear case, if two objects exert
torques on each other, the total angular momentum will be conserved because the torques are interior to the system as a whole.
As the guy collides with the ball, he pushes off with his feet. He applies a torque to the ball, thus speeding up its rotation rate. Not to be outdone, the ball applies a torque back on the guy,
causing him to spin in the opposite direction. The result of all of this: the ball speeds up, the guy slows down, and he starts spinning in the opposite direction to the ball. Linear and angular
momentum are conserved, he sticks the landing, and everyone is happy! Now
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Matches for: Author/Editor=(Sarich_Marco)
Courant Lecture Notes
2013; 128 pp; softcover
Volume: 24
ISBN-10: 0-8218-4359-1
ISBN-13: 978-0-8218-4359-8
List Price: US$34
Member Price: US$27.20
Order Code: CLN/24
Applications in modern biotechnology and molecular medicine often require simulation of biomolecular systems in atomic representation with immense length and timescales that are far beyond the
capacity of computer power currently available. As a consequence, there is an increasing need for reduced models that describe the relevant dynamical properties while at the same time being less
complex. In this book the authors exploit the existence of metastable sets for constructing such a reduced molecular dynamics model, the so-called Markov state model (MSM), with good approximation
properties on the long timescales.
With its many examples and illustrations, this book is addressed to graduate students, mathematicians, and practical computational scientists wanting an overview of the mathematical background for
the ever-increasing research activity on how to construct MSMs for very different molecular systems ranging from peptides to proteins, from RNA to DNA, and via molecular sensors to molecular
aggregation. This book bridges the gap between mathematical research on molecular dynamics and its practical use for realistic molecular systems by providing readers with tools for performing
in-depth analysis of simulation and data-analysis methods.
Titles in this series are co-published with the Courant Institute of Mathematical Sciences at New York University.
Graduate students and research mathematicians interested in molecular dynamics.
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This book uses a consistent notation for its equations. We will describe the symbols we use here, as well as other ways that those concepts are notated in other books.
$abla$: Gradient operator
$abla$^2: Laplacian operator
$\Box$: d'Alembertian operator
A: Area
A: Magnetic potential field
B: Magnetic field
c: Speed of light
E: Electric Field
F[e]: Electric force vector
F[m]: Magnetic force vector
F: Field Tensor
G: Magnetic Field Tensor
I: Current, in amperes
j: Electric current
j: Electric current density
k: Coulomb's Constant
K: Magnetic Constant
l: length
L: Lorentz transformation matrix
M: Magnetization field
n: Normal vector
φ: Electric potential function
r: Scalar distance
r, θ, z: Cylindrical coordinates
r, θ, ρ: Spherical coordinates
r: Distance vector
R: Rotation matrix
σ: Charge density
T[E]: Electric stress tensor
T[M]: Magnetic stress tensor
V: Volume
x, y, z: Cartesian Coordinates
Last modified on 5 July 2011, at 16:42
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Would mathematics teaching be considered "higher level maths discussion"?
May 8th 2011, 10:17 PM #1
Would mathematics teaching be considered "higher level maths discussion"?
As the title suggests, this is to ask a question whether you would consider discussion about mathematics teaching to fit into this subforum?
And as a treat - a fun video I've come across which could be our first discussion...
The Escapist : Video Galleries : Extra Credits : Gamifying Education
I like the ideas, and I can definitely see applications to mathematics education (you could have students work out their own EXP for example...) Thoughts?
As the title suggests, this is to ask a question whether you would consider discussion about mathematics teaching to fit into this subforum?
And as a treat - a fun video I've come across which could be our first discussion...
The Escapist : Video Galleries : Extra Credits : Gamifying Education
I like the ideas, and I can definitely see applications to mathematics education (you could have students work out their own EXP for example...) Thoughts?
Theoretically something under the 'teaching category' would fit nicely into this subforum, but this particular post does not. I believe that it is more of a chat room post.
Well I would like to discuss the ideas from the video I posted, which would fit very nicely into this subforum then
Yeah, I think that's how we'll go about it.
I think it will be best to moderate this forum as little as possible so as to encourage discussion, even if it turns out not being "advanced".
May 8th 2011, 10:44 PM #2
May 8th 2011, 10:49 PM #3
May 8th 2011, 10:49 PM #4
May 9th 2011, 11:52 AM #5
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Velocity Reviews - Locally static?!
> constant B4 : unsigned := "010";
> constant CONST : unsigned := B4 + "001"; --is NOT locally static
> constant B4 : integer := 2;
> constant CONST : integer := B4 + 1; -- is locally static
From the LRM:
7.4.1 Locally static primaries
An expression is said to be locally static if and only if every
operator in the expression denotes an implicitly defined operator
whose operands and result are scalar and if every primary in the
expression is a locally static primary, where a locally static
primary is defined to be one of the following: ...
Integers are scalar, arrays are not.
The VHDL-200X effort has proposals that modify this.
Two important changes:
1) Strike the requirement for the operand and result to be scalar. (FT22)
2) Include operators defined in std_logic_1164 and numeric_std along
with implicitly defined operators in consideration for being
locally static. (FT23)
The VHDL-200X working group is looking for corporate support
to fund the LRM editing effort. Funding will determine the
question as to when the features will be available.
If your company can help sponsor this effort, please contact
myself (VASG vice-chair) or Stephen Bailey (VASG chair).
Best Regards,
Jim Lewis
> Taras
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~
Jim Lewis
Director of Training mailto:Jim@SynthWorks.com
SynthWorks Design Inc.
Expert VHDL Training for Hardware Design and Verification
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~
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Stat Key
Standings Individuals Scoreboard Team Detail Individual Detail Stat Key
Stat Key
│W, L, T│Number of games won (W), lost (L), or tied (T) │
│Pct │Fraction of games won │
│PPG │Average number of points scored by the team in a game │
│PAPG │Average number of points scored against the team in a game │
│Mrg │Team's average margin of victory (if positive) or defeat (if negative) │
│10 -5 │Number of toss-ups answered for the corresponding point value (negative point values are for incorrect interrupts) │
│TUH │Total number of toss-ups heard by the team │
│PPTH │Average number of points scored by the team per toss-up heard │
│P/N │Ratio of powers to negs │
│BHrd │Total number of bonus questions heard by the team │
│BPts │Total number of points scored by the team on bonus questions │
│P/B │Average number of points scored by the team on bonus questions │
│GP │Number of games in which the player participated │
│10 -5│Number of toss-ups answered for the corresponding point value (negative point values are for incorrect interrupts) │
│TUH │Total number of toss-ups heard by the player │
│P/TU │Average number of points scored by the player per toss-up heard │
│P/N │Ratio of powers to negs │
│Pts │Total number of points scored by the player │
│PPG │Average number of points scored by the player per game │
For each game, the score is listed in large bold print on the first line. Then each individual who played in the game is listed, by team, along with the number of each type of question answered (in
the order that they appear in the other reports (usually decreasing order, such as 15 10 -5). The last number after each name is the individual's total points for the game. The next line of the
boxscore gives, for each team, the total number of bonuses heard, the total number of points scored on bonuses, and the average number of points scored per bonus heard.
│Result│Whether the team won (W), lost (L), or tied (T) the game │
│PF │Total number of points scored by the team │
│PA │Total number of points scored by the opponent │
│10 -5 │Number of toss-ups answered for the corresponding point value (negative point values are for incorrect interrupts │
│TUH │Total number of toss-ups heard │
│PPTH │Average number of points scored by the team per toss-up heard │
│P/N │Ratio of powers to negs │
│BHrd │Total number of bonus questions heard by the team │
│BPts │Total number of points scored by the team on bonus questions │
│P/B │Average number of points scored by the team on bonus questions │
│GP │Number of games in which the player participated │
│10 -5│Number of toss-ups answered for the corresponding point value (negative point values are for incorrect interrupts) │
│TUH │Total number of toss-ups heard by the player │
│P/TU │Average number of points scored by the player per toss-up heard │
│P/N │Ratio of powers to negs │
│Pts │Total number of points scored by the player │
│PPG │Average number of points scored by the player per game │
│PPG/Team │Average number of points scored per team per game │
│TUPts/TUH.│Average number of points scored on toss-up questions per toss-up heard │
│BPts/BH │Average number of points scored on bonus questions per bonus heard │
For information about this statistics program, visit the SQBS homepage.
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[FOM] When is it appropriate to treat isomorphism as identity?
Arnon Avron aa at tau.ac.il
Fri May 8 14:03:48 EDT 2009
This is a somewhat late addition to the discussion
when is it appropriate to take isomorphism as identity,
caused by coming across another example where
(in my opinion) such an identification is counter-intuitive:
It is not difficult (see Fitting's new book "incompleteness
in the land of sets") to define a p.r. relation E on N
(the natural numbers) so that <N,E> is isomorphic
to the structure <HF,\in> (where HF is the set of hereditarily
finite sets, and \in is the "epsilon" relation). So
presumably <HF,\in> and <N,E> are "identical". On
the other hand <N,E> and the usual standard structure
<N,0,S,+,*> are identical in the same sense that
<N,0,S,+,*> and <N,0,<,S,+,*,> are identical: E is definable
in terms of 0,S,+,*, and vice versa.
So the conclusion should be that <HF,\in> and <N,0,S,+,*>
are "identical". But are they really? I do not think so
(to start with: it is not a trivial mathematical fact
that these two structures are "identical", but
how can one even *formulates* this nontrivial mathematical fact
without distinguishing first between the two structures?).
Another famous, very common identification which
seems to me problematic is that of the Euclidean plane
with R^2 (where R is the "set" of "real" numbers).
The concept of a "real number" was historically
problematic, and in my opinion it still is, since it
depends on the very questionable powerset axiom.
On the other hand (And I know that I am in a minority
here, in the present cultural climate), I agree with Kant
and Frege that we have a direct intuition concerning
the meaning and TRUTH of the theorems of the Euclidean
Geometry. I was never able to persuade myself otherwise
(and the many beautiful pictures I have seen of the
"Mandelbrot set" tell me that so do many mathematicians
and other scientists, at least in some days of the week).
Arnon Avron
More information about the FOM mailing list
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Analog Signal Correlating Using an Analog-Based Signal Conditioning Front End
Issue Archive
Analog Signal Correlating Using an Analog-Based Signal Conditioning Front End
Friday, 01 March 2013
Converting a signal into a bit stream simplifies the correlation function calculation.
This innovation is capable of correlating two analog signals by using an analog- based signal conditioning front end to hard-limit the analog signals through adaptive thresholding into a binary bit
stream, then performing the correlation using a Hamming “similarity” calculator function embedded in a one-bit digital correlator (OBDC). By converting the analog signal into a bit stream, the
calculation of the correlation function is simplified, and less hardware resources are needed. This binary representation allows the hardware to move from a DSP where instructions are performed
serially, into digital logic where calculations can be performed in parallel, greatly speeding up calculations.
Each of two analog signals (channels A and B) is converted to a digital bit stream by phase correcting it and comparing it to an average of itself at a sampling clock rate f. The hard-limited
conversions of A and B are bitwise compared to measure the level of similarity between the two by the OBDC. This similarity measurement X is equal to the maximum possible Hamming distance (N bits in
disagreement) minus the measured number of bits in disagreement.
The OBDC functions are embedded into a field programmable gate array (FPGA). The OBDC is made up of two shift registers containing the current sample values (of length N) from each of the two input
channels (A and B). During each sample clock, a new sample from each A and B input is clocked into the input linear shift register for each respective channel; this input shifts the current values in
the linear shift register. The oldest (N + 1 sample clocks ago) sample is clocked out of the register. Once the inputs have been clocked in, the correlation routine can start. This rising edge of the
sample clock also clears the max correlation value, the best correlation index, and the shift counter registers, initializing the correlator.
When the two registers match exactly, or are correlated, the X value will equal N. Once the correlation value has been calculated, this result is forwarded to compare with the max correlation value
register. If the X value is greater than the current max correlation value, then the max correlation value becomes X, and the shift counter register is latched and put into the best correlation index
register, providing the index of the current best correlation.
This index is the number of sample clock periods difference between the two input signals and thus, for sample clock rate f, indicates the delay between the signals A and B.
This work was done by Norman Prokop and Michael Krasowski of Glenn Research Center.
Inquiries concerning rights for the commercial use of this invention should be addressed to NASA Glenn Research Center, Innovative Partnerships Office, Attn: Steven Fedor, Mail Stop 4–8, 21000
Brookpark Road, Cleveland, Ohio 44135. LEW-18902-1
This Brief includes a Technical Support Package (TSP).
Analog Signal Correlating Using an Analog-Based Signal Conditioning Front End (reference LEW-18902-1) is currently available for download from the TSP library.
Please Login at the top of the page to download.
Digital Edition
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US GLOBEC: Frontal Exchange Processes over Eastern Georges Bank
Moored EF Toroid buoy Data 1999 - Hourly
eftsc1m_1hr.mat is the EF toroid Seacat Matlab file at a depth of 1-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
eftvm5m_1hr.mat is the EF toroid VMCM Matlab file at a depth of 5-m sampling V/T every hour.
Variables jday (julian day), temp (temperature C) and velocity (cm/s)
eftbt10m_1hr.mat is the EF toroid T recorder Matlab file at a depth of 10-m sampling T every hour.
Variables jday (julian day), temp (temperature C)
eftsc15m_1hr.mat is the EF toroid Seacat Matlab file at a depth of 15-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
eftbt20m_1hr.mat is the EF toroid T recorder Matlab file at a depth of 20-m sampling T every hour.
Variables jday (julian day), temp (temperature C)
eftvm25m_1hr.mat is the EF toroid VMCM Matlab file at a depth of 25-m sampling V/T every hour.
Variables jday (julian day), temp (temperature C) and velocity (cm/s)
eftbt45m_1hr.mat is the EF toroid T recorder Matlab file at a depth of 45-m sampling T every hour.
Variables jday (julian day), temp (temperature C)
eftsc80m_1hr.mat is the EF toroid Seacat Matlab file at a depth of 80-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
Moored EF Mooring Discus buoy Data 1999 - hourly
efdsc2m_1hr.mat is the EF discus Seacat Matlab file at a depth of 2-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
efdvm5m_1hr.mat is the EF discus VMCM Matlab file at a depth of 5-m sampling V/T every hour.
Variables jday (julian day), temp (temperature C) and velocity (cm/s)
efdbt10m_1hr.mat is the EF discus T recorder Matlab file at a depth of 10-m sampling T every hour.
Variables jday (julian day), temp (temperature C)
efdsc15m_1hr.mat is the EF discus Seacat Matlab file at a depth of 15-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
efdbt20m_1hr.mat is the EF discus T recorder Matlab file at a depth of 20-m sampling T every hour
Variables jday (julian day), temp (temperature C)
eftvm25m_1hr.mat is the EF discus VMCM Matlab file at a depth of 25-m sampling V/T every hour.
Variables jday (julian day), temp (temperature C) and velocity (cm/s)
efdsc35m_1hr.mat is the EF discus Seacat Matlab file at a depth of 35-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
efdbt45m_1hr.mat is the EF discus T recorder Matlab file at a depth of 45-m sampling T every hour.
Variables jday (julian day), temp (temperature C)
efdbt60m_1hr.mat is the EF discus T recorder Matlab file at a depth of 60-m sampling T every hour.
Variables jday (julian day), temp (temperature C)
efdsc80m_1hr.mat is the EF discus Seacat Matlab file at a depth of 80-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
Moored EF Guard buoy Data 1999 - hourly
efgsc1m_1hr.mat is the EF guard Seacat Matlab file at a depth of 80-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
Moored NFS toroid/discus buoy Data 1999 - hourly
nfsssc2m_1hr.mat is the NFS Seacat Matlab file at a depth of 2-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfsscm5m_1hr.mat is the NFS VMCM Matlab file at a depth of 5-m sampling V/T every hour.
Variables jday (julian day), temp (temperature C) and velocity (cm/s)
nfsssc10m_1hr.mat is the NFS Seacat Matlab file at a depth of 10-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfsscm15m_1hr.mat is the NFS VMCM Matlab file at a depth of 15-m sampling V/T every hour.
Variables jday (julian day), temp (temperature C) and velocity (cm/s)
nfsssc20m_1hr.mat is the NFS Seacat Matlab file at a depth of 20-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
Moored NFS subsurface buoy Data 1999 - hourly
nfssvm24m_1hr.mat is the NFS VMCM Matlab file at a depth of 24-m sampling V/T every hour.
Variables jday (julian day), temp (temperature C) and velocity (cm/s)
nfsssc34m_1hr.mat is the NFS Seacat Matlab file at a depth of 34-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfssvm44m_1hr.mat is the NFS VMCM Matlab file at a depth of 44-m sampling V/T every hour.
Variables jday (julian day), temp (temperature C) and velocity (cm/s)
nfsssc54m_1hr.mat is the NFS Seacat Matlab file at a depth of 54-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfssvm64m_1hr.mat is the NFS VMCM Matlab file at a depth of 64-m sampling V/T every hour.
Variables jday (julian day), temp (temperature C) and velocity (cm/s)
nfssssg72m_1hr.mat is the NFS Seacat Matlab file at a depth of 72-m sampling T/C/S/P every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
Moored NFD toroid/discus buoy Data 1999 - hourly
nfdtsc2m_1hr.mat is the NFD toroid Matlab file at a depth of 2-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfdtvm5m_1hr.mat is the NFD toroid VMCM Matlab file at a depth of 5-m sampling V/T every hour.
Variables jday (julian day), temp (temperature C) and velocity (cm/s)
nfdtsc15m_1hr.mat is the NFD toroid Matlab file at a depth of 15-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfdtsc30m_1hr.mat is the NFD toroid Matlab file at a depth of 30-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfdtsc50m_1hr.mat is the NFD toroid Matlab file at a depth of 50-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfdtsc75m_1hr.mat is the NFD toroid Matlab file at a depth of 75-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfdtsc100mhour.mat is the NFD toroid Matlab file at a depth of 10-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfddsc2m_1hr.mat is the NFD discus Matlab file at a depth of 2-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfddvm5m_1hr.mat is the NFD discus VMCM Matlab file at a depth of 5-m sampling V/T every hour.
Variables jday (julian day), temp (temperature C) and velocity (cm/s)
nfddsc30m_1hr.mat is the NFD discus Matlab file at a depth of 30-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfddsc50m_1hr.mat is the NFD discus Matlab file at a depth of 50-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
nfddsc75m_1hr.mat is the NFD discus Matlab file at a depth of 75-m sampling T/C/S every hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
Moored NEP surface buoy Data 1999 - hourly
neptc1m_1hr.mat is the NEP mooring Matlab file at a depth of 1-m sampling T/C/S every 3hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
neptc5m_1hr.mat is the NEP mooring Matlab file at a depth of 5-m sampling T/C/S every 3hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
neptc15m_1hr.mat is the NEP mooring Matlab file at a depth of 15-m sampling T/C/S every 3hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
neptc50m_1hr.mat is the NEP mooring Matlab file at a depth of 50-m sampling T/C/S every 3hour.
Variables jday (julian day), temp (temperature C) and salt (salinity ppt)
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A spectre is haunting Europe
-- Karl Marx and Friedrich Engels (1848)
During the Renaissance, when people looked through prisms at each other, they saw what appeared to be ghosts or spectra, multicolored and transparent images floating mysteriously near
each other. What they were seeing, of course, were the colors of the image separating due to optical dispersion of the glass. The first scientific use of this phenomena was made by
Newton, whose own theory of spectra turned out to be one of the few mistakes he published in physics. The true explanation was to wait for the contributions of W. H. Wollaston and J.
Fraunhofer in the early 1800s, who based their results on the assumption that light was a wave, not a particle. (We now know it is both but Newton was still wrong.) They improved their
experiments by using a slit instead of a pinhole to collimate their light. This produced a line rather than a spot, and as a legacy we still refer to spectral lines, even when the data
are handled digitally, and the frequency structure shown as a peak in a graph rather than as a brightly colored line on a white sheet. A group of spectral peaks is still called a "band,"
which is what a group of adjacent lines looked like. The band-width was the width in the optical spectrum of a group of lines; we still use this term today (without the hyphen) to denote
the range of frequencies covered by a system.
Today, spectral analysis refers to any process that accounts for data as a sum of individual oscillations. When a musician listens to a chord played on a piano, and then goes to the piano
and duplicates it, he has successfully performed spectral analysis. He has taken a complex sound and recognized it as a sum of individual oscillations. This is done using the hair cells
of the inner ear, which are separated in the cochlear duct by their frequency sensitivity. That is what spectral analysis is separation of a signal into different frequency components.
The concept is made more complicated by the fact that every sound, even the roar of a waterfall, can be constructed from pure tones, although it takes a lot of them. But it is easier to
reproduce the sound of a waterfall by recreating the amplitude of the sound as a function of time. This is called working in the "time domain." On the other hand, when we analyze a signal
by breaking it up into a sum of pure frequencies, we say we are working in the "frequency domain."
Paleoclimate signals usually consist of a superposition of a few pure frequencies (also called lines, or tones) and background. Some of the analysis is best done in the time domain. This
includes the precise timing of the glacial terminations, and the history of ice-rafting phenomena known as Heinrich events. A sudden event in the time domain, when described in the
frequency domain, appears spread out over many frequencies, so it is difficult to study that way. Likewise, a single peak appearing in the frequency domain (e.g. at 41 kyr), appears as a
relatively complex object in the time domain: a sinusoidal variation that covers all time. Spectral analysis and time domain analysis are best done together, since certain phenomena stand
out in one representation, and other in the other. Non-periodic events are just as interesting as periodic ones; after all, it was a single dramatic event that appears to have killed the
dinosaurs. But the frequency domain is extremely useful for many other astronomical forces (such as insolation) because astronomical signals are often extremely regular in their
repetitiveness. They tend to stay in tune for a long time, and that gives rise to very strong and narrow peaks in the frequency domain.
There is interesting physics behind the fact that the astronomical signals are so regular. The first is the remarkable fact that the orbits of the planets happen to be nearly perfect
ellipses, as mentioned earlier. This arises because the force of gravity is nearly a perfect 1/r^2 force. It didn t have to be this way. If the force of gravity were 1/r, or 1/Ãr, or 1/r^
3, we would not have closed orbits. In fact, the only other force law that gives a closed orbit is the linear spring law, F = kr. (See L. D. Landau and E. M. Lifshitz, Mechanics, 3^rd
edition, p. 32.) But because gravity varies with the inverse square of the separation, the orbits repeat. If friction were present (and it is for really small particles; see Section 7.2),
then the orbits would not close. General relativity theory is also a slight departure from the inverse square law, and a consequence of this is that the orbits of the planets do not
completely close. One consequence is the "advance of the perihelion of Mercury." When astronomers measure this, they are studying the departure from the Newtonian inverse-square law.
Perturbations from the other planets also cause departures from pure single-frequency periodicity, for the same reason. Although they are individually inverse-square, each planet s
contribution to the force comes from a different direction in space than that of the Sun, so when added to the Sun s gravity, the net resulting force is not inverse square.
The most natural way to do spectral analysis of data is to try to fit it to a mathematical function containing sine waves. For example, if we have data points y[k], each having an age t
[k] then we could try to find the best parameters a[0], a[k], f[k], and f[k], such that we minimize the difference between the real data and the function:
is minimized. (For an illustration of this processes, you can look ahead to our description of eccentricity in Section 2.2.2; in particular, Figure 2-7 on page * shows an orbital function
called eccentricity closely approximated by the sum of three sine waves.) Many computer programs have been written for just this minimization process. (In the Matlab™ language, for example,
the program is called fmins.m.) This approach is excellent, in principle; it is closely related to a method we will describe (Section 3.7) as the "Maximum Likelihood Method." Its main
disadvantage is that it is very time consuming. A much faster method is to use the Fourier transform, for which very rapid methods of computation have been devised. But if part of the signal
is periodic, and part is background, and if the time scale is uncertain, then performing a Fourier transform is often not the optimum approach. The discipline called spectral analysis has
developed largely to determine the best approach to analyzing a periodic signal in the presence of background and uncertainty. The choice of technique to use depends on the nature of the
signal (is it broad? a sine wave? a square wave?), the nature of the noise (Gaussian? broad or narrow band?), the accuracy of the data (error in time scale?), and the nature of the variable
you want to measure (amplitude of the periodic component? frequency? phase? spectral shape?). Not only is such optimization important for science, but it can be extremely important in a
competitive industry such as communications. The result is that the field of spectral analysis has become very sophisticated, and very arcane. Sometimes it feels too arcane so full of
traps, and so difficult to master, that otherwise excellent scientists just try to avoid it.
To make the optimum choice of spectral analysis technique requires a detailed understanding of the background, i.e. the part of the signal that we don t want to study. But we usually don t
have that. The result is that there is no "correct" way to do the analysis, only ways that have different advantages and disadvantages, and different traps. The field of paleoclimate has been
dominated for a long time by the use of the "Blackman-Tukey method," presented in detail in the classic 1958 book The measurement of power spectra . It is also reportedly the method that
Tukey himself suggested be used for paleoclimate work. It is, in fact, particularly well suited for data with an uncertain time scale, as has been the case for much paleoclimate data.
We ll give a simple example of spectral analysis here, while leaving the details to a later chapter. In an Antarctic ice core from Vostok, the relative fraction of the hydrogen isotope
deuterium was measured as a function of depth. Using a model for the ice flow, this could be plotted as a function of time. The data were part of the basis for Figure 1-5. We show these data
covering the entire period 0 to 420 kyr, in Figure 1-8.
There appears to be a periodicity of approximately 100 kyr. This is evident in the spectral power plot shown in Figure 1-9. The plot was made using the periodogram method of Section 3.2.4.
Figure 1-9 Spectrum of Vostok deuterium
From this plot, we see that in addition to the 100 kyr cycle (the peak near the frequency f = 0.01 cycles/ky), there is another prominent peak near 0.024 cycles/kyr. This is identified with
an orbital parameter called obliquity, which is described in Section 2.2.4. It is not evident in the time domain because the eye is distracted by the strong 100 kyr cycle.
Our real purpose in showing the spectra here is to demonstrate how phenomena that are well hidden in the time domain, can stand out in the frequency domain. And it works both ways. The
"sudden terminations" in the time plot are the nearly vertical lines that appear at the end of each glaciation. There is no evidence for these in the spectral plot. Such sharp changes require
a conspiracy of a large number of small components at many different frequencies. These small contributions are well-hidden in the frequency domain. A comprehensive analysis of data requires
analysis in both the time and frequency domains. Often it is necessary to do both, by breaking the data into segments of different time intervals, and doing the spectral analysis on each
segment. This is important when the dominant cycles change with time, as they often do in paleoclimate data.
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N-Way ANOVA example Two-way analysis of variance is where the rubber hits the road, so to speak. This extends the concepts of ANOVA with only one factor to two factors. When there are two factors
this means that there can be an interaction between the two factors that should be tested. As one might expect
A quicky..
If you’re (and you should) interested in principal components then take a good look at this. The linked post will take you by hand to do everything from scratch. If you’re not in the mood then the
dollowing R functions will help you. An example. # Generates sample matrix of five discrete clusters that have
One-way Analysis of Variance (ANOVA)
Analysis of Variance (ANOVA) is a commonly used statistical technique for investigating data by comparing the means of subsets of the data. The base case is the one-way ANOVA which is an extension of
two-sample t test for independent groups covering situations where there are more than two groups being compared. In one-way ANOVA the data
Create factor variables in R
Instead of the factor() function which usually applies after defining a vector there’s the gl() base function to do this in one step, egfreq <- c(204,6,1,211,13,5,357,44,38,92,34,49) row <- gl
(4,3,length=12) col <- gl(3,1,length=12) > col 1 2 3 1 2 3 1 2 3 1 2 3 Levels: 1 2 3 tt <- data.frame(freq,row,col) > xtabs(tt) col row 1 2 3 1 204 6
Design of Experiments – Optimal Designs
When designing an experiment it is not always possible to generate a regular, balanced design such as a full or fractional factorial design plan. There are usually restrictions of the total number of
experiments that can be undertaken or constraints on the factor settings both individually or in combination with each other. In these scenarios computer
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the graded pieces of the gamma-filtration of Quillen K-theory and Chow groups of a regular scheme
up vote 3 down vote favorite
Let $X$ be a regular scheme and consider Grothendieck's $\gamma$-filtration $F^nK(X)$ on $K(X)$. For the graded pieces, one has $Gr^0K(X) = CH^0(X)$ and $Gr^1K(X) = \mathrm{Pic}(X) = CH^1(X)$. Does
this continue to hold, i.e., do we have $Gr^pK(X) = CH^p(X)$?
I found that for $X/k$ smooth quasi-projective, $CH^q(X,p) \otimes \mathbf{Q} = K_p(X)^{(q)} \otimes \mathbf{Q}$, so this holds after rationalising.
ag.algebraic-geometry motivic-cohomology kt.k-theory-homology algebraic-k-theory
1 Note that I've rewritten the second paragraph of my answer, which originally contained too strong a statement. – Steven Landsburg Jun 10 '13 at 19:53
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1 Answer
active oldest votes
The map between the graded $K$-theory ring on the one hand and the Chow ring on the other is defined via Chern classes and requires denominators. I know of no good reason to expect an
integral isomorphism (or even a map), but I'm not aware of an explicit counterexample (though I'm vaguely aware that the experts think the place to look for that counterexample is
over a field with large etale cohomological dimension).
up vote 4 down
vote accepted On the other hand, if you replace the $\gamma$-filtration with the filtration by codimension of support, then you do get $Gr^p(X)$ as a quotient (with torsion kernel) of $Ch^p(X)=H^p
(X,K_p)$ (over the integers) provided $X$ is both regular and of finite type over a field --- though it would follow from Gersten's conjecture that this holds for all regular $X$.
Thank you. By the way, one has $F^1\gamma K(X)=F^1_{top}K(X)$ if there is an ample sheaf. – Timo Keller Jun 10 '13 at 16:07
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Not the answer you're looking for? Browse other questions tagged ag.algebraic-geometry motivic-cohomology kt.k-theory-homology algebraic-k-theory or ask your own question.
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How long do I have to walk to lose 1 pound (lb)?
It can take 5-20 hours of walking to lose 1 pound.
It depends on how much you weigh and how fast you walk. Below is an estimate of how long you’ll have to walk (at various paces) to burn 3500 calories, the number of calories in one pound of fat.
2.0 mile per hour (mph) pace (30 minute mile)
• If you weigh 150 lbs, you’ll have to walk for approximately 20 hours to lose 1 lb.
• If you weigh 200 lbs, you’ll have to walk for approximately 15 hours to lose 1 lb.
• If you weigh 250 lbs, you’ll have to walk for approximately 12 hours to lose 1 lb.
• If you weigh 300 lbs, you’ll have to walk for approximately 10 hours to lose 1 lb.
• If you weigh 350 lbs, you’ll have to walk for approximately 9 hours to lose 1 lb.
• If you weigh 400 lbs, you’ll have to walk for approximately 8 hours to lose 1 lb.
2.5 mph pace (24 minute mile)
• If you weigh 150 lbs, you’ll have to walk for approximately 16 hours to lose 1 lb.
• If you weigh 200 lbs, you’ll have to walk for approximately 12 hours to lose 1 lb.
• If you weigh 250 lbs, you’ll have to walk for approximately 10 hours to lose 1 lb.
• If you weigh 300 lbs, you’ll have to walk for approximately 8 hours to lose 1 lb.
• If you weigh 350 lbs, you’ll have to walk for approximately 7 hours to lose 1 lb.
• If you weigh 400 lbs, you will have to walk approximately 6 hours to lose 1 lb.
3.0 mph pace (20 minute mile)
• If you weigh 150 lbs, you’ll have to walk for approximately 15 hours to lose 1 lb.
• If you weigh 200 lbs, you’ll have to walk for approximately 11 hours to lose 1 lb.
• If you weigh 250 lbs, you’ll have to walk for approximately 9 hours to lose 1 lb.
• If you weigh 300 lbs, you’ll have to walk for approximately 7 hours to lose 1 lb.
• If you weigh 350 lbs, you’ll have to walk for approximately 6 hours to lose 1 lb.
• If you weigh 400 lbs, you’ll have to walk for approximately 6 hours to lose 1 lb.
3.5 mph pace (17 minute mile)
• If you weigh 150 lbs, you’ll have to walk for approximately 13 hours to lose 1 lb.
• If you weigh 200 lbs, you’ll have to walk for approximately 10 hours to lose 1 lb.
• If you weigh 250 lbs, you’ll have to walk for approximately 9 hours to lose 1 lb.
• If you weigh 300 lbs, you’ll have to walk for approximately 6 hours to lose 1 lb.
• If you weigh 350 lbs, you’ll have to walk for approximately 5 hours to lose 1 lb.
• If you weigh 400 lbs, you’ll have to walk for approximately 5 hours to lose 1 lb.
Ainsworth BE, Haskell WL, Whitt MC, et al. Compendium of physical activities: an update of activity codes and MET intensities. Med. Sci. Sports Exerc., Vol. 32, No. 9, Suppl., pp. S498–S516, 2000.
Calorie calculations from equation: (METs x 3.5 x body weight in kg)/200 = calories/minute
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lebesgue measure and differentiability
January 14th 2013, 12:22 PM #1
Junior Member
Oct 2012
lebesgue measure and differentiability
Hi there
I've got the following two exercises to solve:
Consider the measure space $\([0,\infty\),B,\lambda\)$ where B denotes the Borel-sigma-algebra and $\lambda$ the Lebesgue-measure.
Let $\mu << \lambda$ and $\frac{d\mu}{d\lambda}$ continuous. Show that
$f(x)=\mu([0,x])$ is differentiable and that
If $\frac{d\mu}{d\lambda}$ is $\lambda-$ almost everywhere continuous only is then
$f'(x)=\frac{d\mu}{d\lambda}(x)$$\lambda-$almost everywhere as well?
Why? Prove your answer!
According to the theorem of Radon-Nikodym for $g(x)=\frac{d\mu}{d\lambda}(x)$ then
$f(x)= \int_{[0,x]} g d\lambda(x)$
Obviously the derivative of this is g but how can one show this in a proper way?
$f'(x)= \int_{[0,x]} g \frac{d\lambda(x)}{d\lambda(x)} =\int_{[0,x]} g 1$????
I also tried it this way:
$f'(x)=\frac{d\mu}{d\lambda}(x)-\frac{d\mu}{d\lambda}(0)$ but what then is x and 0? Definitely not a one-point set... cause then it was all zero.
I'd say that this is right because where
$f'(x) eq \frac{d\mu}{d\lambda}$ is a Lebesgue-zero set but how to show?
Could please someone help me to solve this exercise? Any hint would be appreciated.
Re: lebesgue measure and differentiability
Re: lebesgue measure and differentiability
Thank you very much.
Concerning 2)
A kind of "prove" or argument hmmm
assuming g is continuous at $x_0$ and $x_0$ is contained in a non-zero-set of $\lambda$ then this proof works also, ok.
But when x satisfies $|x-x_0|<\delta$ and x is on the left of $x_0$...doesn't that mean that at x the function g is continuous from the right since the $\delta$ is arbitraty ? That's right yeah?
Re: lebesgue measure and differentiability
Hi there
Well I understand the proof but I still don't understand why this is also valid for a.e. continuity. Can someone please explain me that?
Cause then I have
$\int_{[x_0,x_0+h]}|g(x)-g(x_0) | dx$ is 0 because the set $[x_0,x_0+h]$ is a zero set but then I still have this $\frac{1}{h}$ which goes to infinity when h goes to 0.
Is there anyone who can answer me that please?
January 15th 2013, 07:12 PM #2
Super Member
Dec 2012
Athens, OH, USA
January 16th 2013, 04:03 AM #3
Junior Member
Oct 2012
February 1st 2013, 02:24 PM #4
Junior Member
Oct 2012
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Prototaip perisian PBK matematik interaktif (pecahan algebra) menggunakan pendekatan EIF
Ahmad, Norazlina (2003) Prototaip perisian PBK matematik interaktif (pecahan algebra) menggunakan pendekatan EIF. Masters thesis, Universiti Teknologi Malaysia, Faculty of Computer Science and
Information System.
Restricted to Repository staff only
PDF (Abstract)
PDF (Table of Contents)
PDF (1st Chapter)
Mathematics is a subject that requires students to understand the concept of learning in a comprehensive way with forming a scheme on the concept. To achieve this goal, students have to understand
the basic concepts and examples given. Then the students will have to do exercises given and some extra exercises on their own to strengthen their knowledge. In performing this learning process, a
study was done on using the EIF approach in the PBK software prototype, also known as the Computer Assisted Learning - Algebra Fraction (PBK - PA) that contains topics on algebra fraction for
semester one syllabus in Politeknik Malaysia. The EIF approach involves three stages, which are explain (Q, Instruct (I), and Facilitate (F) that must he implemented throughout the process. The PBK-
PA prototype software also focus on the teaching and learning theory, teaching strategy and the use of multimedia technology. This prototype is developed using software as Microsoft Visual Basic 6.0,
Swish, Adobe Photoshop and Sound Forge. The outcome of this project is a PBK - PA software prototype that can help students to carry out the learning process effectively
Repository Staff Only: item control page
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Erich von Daniken's "Chariots of the Gods?":
Erich von Daniken's "Chariots of the Gods?":
Science or Charlatanism?
by Robert Sheaffer
(First published in the "NICAP UFO Investigator", October/November, 1974.)
Was God an ancient astronaut? Do centuries-old legends of gods and heroes tell of space travelers who came to earth from distant parts of the Cosmos? Are some of the ruins of antiquity remnants of
great airfields, the favored landing sites of extraterrestrial craft?
"Yes!," writes Erich Von Daniken in his runaway bestselling book, "Chariots of the Gods?" This phenomenally successful book is now in its forty-fourth printing, with over four million copies
currently in print. Von Daniken's sequels to this book, "Gods from Outer Space" and "The Gold of The Gods" are also selling well, as are flocks of imitations.
The popularity of such a sensational theory should not be surprising. Immanuel Velikovsky created a similar stir almost twenty-five years ago with the publication of his "Worlds in Collision",
suggesting that the present state of the solar system can be explained by a series of spectacular cataclysms among the planets. It has been over twenty-seven years since "flying saucers" burst into
the public's awareness, and UFOs still continue to generate excitement and controversy. Sensational hypotheses such as these generate such levels of interest that they tend to become self-sustaining,
quite apart from the question of whether they are true.
Established science has always shied away from such remarkable claims. "It took courage to write this book. and it will take courage to read it", says Von Daniken. "Even if a reactionary army tries
to dam up this new intellectual flood, a new world must be conquered in the teeth of all the unteachable in the name of truth and reality." (Presumably the reader here is expected to shout, "Right
on!") Does science avoid Von Daniken because it is afraid to face up to the truth? Let's examine some of his many claims, to see if they are serious scientific theories, or mere humbug.
On page 9, Von Daniken serves up a "basic rocket equation," derived by one Professor Ackeret, purporting to show how time slows down for space travelers who zip along at velocities near the speed of
light. This is an important consequence of Einstein's theory of relativity. Yet a quick glance at this "rocket equation" shows that it isn't an equation at all! Every equation is a mathematical
statement of the equality of two quantities: this equals that. But his "equation" contains no equal sign, and hence it cannot be a real equation; it must be intended as window dressing, since it
serves no legitimate mathematical purpose. Yet that is not the only absurdity in this non-equation. A term in the denominator is multiplied by a very strange constant: ONE! Did Professor Ackeret
think that multiplication by one was a necessary step in his calculations? Von Daniken himself must have known better, as even schoolchildren learn that any number times one equals itself! There are,
of course, many legitimate equations which deal with this aspect of the relativity of time. Why then has Von Daniken selected such an obvious humbug to support his "scientific " claims?
What is the history of our earth-moon system? ". . . a satellite was captured by the earth. As it was pulled toward the earth it slowed down the speed of the earth's revolutions. It finally
disintegrated and was replaced by the moon." (p. 19)
Proof of this theory, he writes, can be found in the symbols on the Great Idol in the Old Temple at Tiahuanaco, one of his favorite archeological Wonders. Supposedly this message, dating back 27,000
years, tells of how this satellite emeritus made 425 revolutions around the earth a year, which was then only 288 days.
If the earth formerly circled the sun every 288 days, Kepler's third law implies that the earth must have been at that time much closer to the sun, almost where Venus is now. Are we expected to
believe that during the great Ice Ages the earth was some twenty million miles closer to the sun than it is today? On the other hand, if the year remained unchanged but each day is shorter, we are
faced with another difficulty: since the year is now 365 days, the earth's rotation is faster than in 25,000 B.C., not slower as Von Daniken claims. How could a satellite slowly spiraling earthward
pull both earth and satellite farther away from the sun? Where are the fragments of this disintegrated moon, and where did our current moon come from? Von Daniken gives no answers.
On ancient Egyptian astronomy: " ... why a Sirius calendar? . . . If Sirius appeared on the horizon at dawn at the same time as the Nile flood, it was pure coincidence ... this very interest in
Sirius seems rather peculiar because seen from Memphis, Sirius can be observed only in the early dawn just above the horizon when the Nile floods begin." (p.64-65)Reader, beware - we're dealing with
one who knows the art of deception! Sirius, he neglects to mention, is the brightest star in the sky. His claim that Sirius was hardly visible at all to the Egyptians is simply false. Sirius, in
fact, is visible from anywhere on earth except the extreme North Polar region, and observers in Egypt see that star higher in the sky than we do here in the northern United States, where it dominates
the sky on crisp winter evenings.
There is no dark mystery behind the development of the Sirius- based calendar in Egypt. The priests there noticed a simple regularity: each year, when that brilliant star first became visible in the
morning sky, the Nile flood began. Does this not prove that the Egyptians had contact with a race of space travelers?
On the mysteries of the great Pyramid of Cheops: "is it coincidence that the area of the base of the pyramid divided by twice its height gives the celebrated figure Pi = 3.14159?" (p. 77)
Here our slippery trickster has made a claim which is easy enough to understand, but its refutation requires a higher level of mathematical sophistication, which is all the better for letting the
deception go unnoticed! Without going into too much detail, let us observe that the famous number Pi is what is called a "dimensionless constant": it is a pure number, with which no units of measure
are associated. However, the ratio of an area to height is not dimensionless, hence such a ratio cannot yield Pi. By choosing our units carefully, we can obtain the number 3.14159, but the ratio will
not really be Pi, which is independent of any units of measure. If we measure the same pyramid, Von Daniken style, in inches, feet, and yards, we will obtain three different ratios. Choose your own
units, and the ratio can be made to equal any number at all! Von Daniken writes that the pyramid of Cheops "has inspired hundreds of crazy and untenable theories". Not satisfied with this collection,
he has given us one more.
More evidence of the ancient Visitors: "The Mayas were intelligent; they had a highly developed culture. They left behind not only a fabulous calendar but also incredible calculations. They knew the
Venusian year of 584 days. . . " (p.55)
This statement is true-almost. He conveniently forgets to tell us that this 584 day period is not the true Venusian year, which is 225 days. Instead it is the apparent Venusian cycle as seen by an
earth-based observer, which is precisely what we should expect the ancient Mayas to record by simply counting days, without any extraterrestrial insight. On the other hand, if they had recorded the
true Venusian year of 225 days, which implies a knowledge of the Copernican (sun-centered) system of astronomy, that would have been a bit more remarkable.
In these few short pages I have scarcely begun a list of the inaccuracies and half-truths to be found in Chariots of the Gods?. But a thousand-page refutation of a hundred-page book would hardly make
good reading. Sensationalist theories have always attracted more readers than refutations of the same.
Enough of Von Daniken's claims have, however, been examined to reveal his method of operations: to dazzle the reader with a skillful blend of half-truths (as well as quarter and eighth- truths, too).
Looking into his past, (Ref: New York Times Book Review), we should not be too surprised to find that his rather broad criterion of truthfulness has, at times, brought him into conflict with the law.
A court in his native Switzerland found Von Daniken guilty of embezzlement, forgery, and fraud, sentencing him to three and a half years in prison. While operating a Swiss hotel. it seems he
fraudulently obtained money by misrepresenting his financial assets, this experience in deception later proving invaluable in his literary career. It was during this stay as a guest of the Swiss
government that he wrote his second book, Gods from Outer Space, now also a best seller.
Once a liar, however, does not infallibly prove him always a liar. However much this charlatanism may hurt one's credibility, it never destroys it completely. (The followers of famous psychics are
never shaken when their leader is caught cheating: they only cheat on bad days!) Von Daniken's theories of Ancient UFO contactees are indeed no more implausible than their latter- day counterparts.
In fact, we have in support of these theories one bona fide, highly reliable eyewitness who saw these creatures disembark who can now describe it: none other than Erich Von Daniken himself! In an
exclusive interview with the National Enquirer, Von Daniken tells of his experiences in Point Aleph, "a sort of fourth dimension'" where time doesn't exist. He revealed how he can now leave his body
at will, transcending all concepts of space and time. "I know that astronauts visited the earth in ancient times", he confides, because "I was there when the astronauts arrived. And I know they'll be
back." Unfortunately for us, he can't say exactly when, since "time doesn't exist in Point Aleph."
"I even know what will happen after death". We're all ears. "l will become part of this huge never-to-be-destroyed ball of energy that keeps and remembers every last tiny thing that has ever happened
on this planet. Everybody will join me there eventually and at least they'll know then that I was right". Save a place for me right in the middle of that big old ball, Erich, because I'm going to be
one of the hardest to convince.
Chariots of the Gods? is currently published by Bantam Books, New York. Page numbers refer to Bantam edition.
Erich Von Daniken's Genesis, New York Sunday Times Book Review. March 31 , 1974
National Enquirer, March 17, 1974
Go to The UFO Skeptic's Page
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Finding asymptotes+Fundamental identities
May 21st 2009, 04:23 PM
Finding asymptotes+Fundamental identities
Hey, I'm new to calculus, and I can't figure out this problem.(Headbang)
You start with:
y = TAN (2 SIN x) -pi < x < pi
Since: TAN x = (SIN x / COS x),
you can get
y= SIN (2 SIN x) / COS (2 SIN x)
Then, since the asymptote is where the equation is undefined (or equals infinity), we need to find the value of x which causes the denominator to equal 0.
COS (2 SIN x) = 0
The cosine of +pi/2 is 0, so
2 SIN x = +pi/2
SIN x = +pi/4
x = + SIN^-1 pi/4
The problem is that there are two answers, but I can't figure out how to work out the next one, which the book says is
x = + (pi - SIN^-1[pi/4]
Help, plz!
I'd also like a way to type in a pi symbol, if that's possible.(Thinking)
May 22nd 2009, 02:34 AM
If you have an equation $\sin x = C$ then the possible solutions are $x=\arcsin C+2n\pi$ and $x=\pi - \arcsin C+2n\pi$
May 24th 2009, 08:18 AM
Is that just a general rule? My Math book doesn't show it. Also, does it work for just $pi - arcsin C$? Otherwise I don't see how it works. The $2n pi$ doesn't seem to fit at all. If it does,
then thanks a lot.(Bow)
May 24th 2009, 09:05 AM
It's the general solution to an equation like this. It contains all the solutions to the equation, but for your problem the question makes it clear that we are only interested in the domain $x\in
[-\pi,\pi]$, so the only value of $n$ that doesn't take us out from the valid domain is $n=0$.
Hence the two solutions are $x=\pm\arcsin\frac{\pi}{4}$ and $x=\pm\left(\pi-\arcsin\frac{\pi}{4}\right)$
If you think about it, going $2n\pi$ radians in a circle always takes us back to where we started.
May 25th 2009, 11:30 AM
Oh, yeah, I guess I was being pretty dumb about the $2n pi$ thing.(Doh) Thanks again.
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Wavelets and Multiscale Signal Processing
- ACHA , 2001
"... We analyze the convergence and smoothness of certain class of nonlinear subdivision schemes. We study the stability properties of these schemes and apply this analysis to the speci c class based
on ENO and weighted-ENO interpolation techniques. Our interest in these techniques is motivated by their ..."
Cited by 20 (5 self)
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We analyze the convergence and smoothness of certain class of nonlinear subdivision schemes. We study the stability properties of these schemes and apply this analysis to the speci c class based on
ENO and weighted-ENO interpolation techniques. Our interest in these techniques is motivated by their application to signal and image processing.
- SIAM J. Math. Anal , 2007
"... Abstract. Refinable functions and cascade algorithms play a fundamental role in wavelet analysis, which is useful in many applications. In this paper we shall study several properties of
refinable functions, cascade algorithms and wavelets, associated with Hölder continuous masks, in the weighted su ..."
Cited by 12 (11 self)
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Abstract. Refinable functions and cascade algorithms play a fundamental role in wavelet analysis, which is useful in many applications. In this paper we shall study several properties of refinable
functions, cascade algorithms and wavelets, associated with Hölder continuous masks, in the weighted subspaces L2,p,γ(R) of L2(R), where 1 � p � ∞, γ � 0 and f ∈ L2,p,γ(R) means
- IEEE Proceedings Part 1 (Vision, Image and Signal Processing , 1997
"... Wavelet transforms have been one of the important signal processing developments in the last decade, especially for applications such as time-frequency analysis, data compression, segmentation
and vision. Although several efficient implementations of wavelet transforms have been derived, their compu ..."
Cited by 9 (2 self)
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Wavelet transforms have been one of the important signal processing developments in the last decade, especially for applications such as time-frequency analysis, data compression, segmentation and
vision. Although several efficient implementations of wavelet transforms have been derived, their computational burden is still considerable. This paper describes two generic parallel implementations
of wavelet transforms based on the pipeline processor farming methodology which have the potential to achieve real-time performance. Results show that the parallel implementation of the over-sampled
Wavelet Transform achieves virtually linear speedup, while the parallel implementation of the Discrete Wavelet Transform (DWT) also out-performs the sequential version, provided that the filter order
is large. The DWT parallelisation performance improves with increasing data length and filter order while the frequency domain implementation performance is independent of wavelet filter order.
Parallel p...
"... This paper is devoted to an approximation problem for operators in Hilbert space, that appears when one tries to study geometrically the cascade algorithm in wavelet theory. Let H be a Hilbert
space, and let π be a representation of L ∞ (T) on H. Let R be a positive operator in L ∞ (T) such that R ( ..."
Cited by 4 (1 self)
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This paper is devoted to an approximation problem for operators in Hilbert space, that appears when one tries to study geometrically the cascade algorithm in wavelet theory. Let H be a Hilbert space,
and let π be a representation of L ∞ (T) on H. Let R be a positive operator in L ∞ (T) such that R (11) = 11, where 11 denotes the constant function 1. We study operators M on H (bounded, but
noncontractive) such that π (f) M = Mπ ( f ( z 2)) and M ∗ π (f)M = π (R ∗ f), f ∈ L ∞ (T), where the ∗ refers to Hilbert space adjoint. We give a complete orthogonal expansion of H which reduces π
such that M acts as a shift on one part, and the residual part is H (∞) = ⋂ n [MnH], where [MnH] is the closure of the range of Mn. The shift part is present, we show, if and only if ker (M ∗ ) ̸=
{0}. We apply the operator-theoretic results to the refinement operator (or cascade algorithm) from wavelet theory. Using the representation π, we show that, for this wavelet
, 2006
"... Abstract. In this paper, we obtain symmetric C ∞ real-valued tight wavelet frames in L2(R) with compact support and the spectral frame approximation order. Furthermore, we present a family of
symmetric compactly supported C ∞ orthonormal complex wavelets in L2(R). A complete analysis of nonstationar ..."
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Abstract. In this paper, we obtain symmetric C ∞ real-valued tight wavelet frames in L2(R) with compact support and the spectral frame approximation order. Furthermore, we present a family of
symmetric compactly supported C ∞ orthonormal complex wavelets in L2(R). A complete analysis of nonstationary tight wavelet frames and orthonormal wavelet bases in L2(R) is given. 1.
"... Abstract. This paper deals with generalized coiflets designed in [Monzón, Beylkin, Hereman, 1999] and with computing their scaling coefficients, respectively. Such wavelets are useful in
applications where interpolation and linear phase are of importance. We derive alternative definitions and prove ..."
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Abstract. This paper deals with generalized coiflets designed in [Monzón, Beylkin, Hereman, 1999] and with computing their scaling coefficients, respectively. Such wavelets are useful in applications
where interpolation and linear phase are of importance. We derive alternative definitions and prove their equivalence. In all definitions the system with minimal number of equations is proposed, the
third definition enables to eliminate some quadratic conditions occurring in original definition. Moreover, by their construction, one additional free parameter is obtained and we propose how to
choose this parameter to further simplify the arising system. Similar approach was also used in [ Černá, Finěk, 2004c]. For small filter lengths this system can be explicitly solved with algebraic
methods like Gröbner bases. The simple structure of arising polynomials (in one variable) allows to find quickly all possible solutions. In addition to the examples of coiflets of length 18 presented
in [Monzón, Beylkin, Hereman, 1999] we have found another ‘maximal coiflets’.
"... Web service development and usage has shifted from simple information processing services to high-value business services that are crucial to productivity and success. In order to deal with an
increasing risk of unavailability or failure of mission-critical Web services we argue the need for advance ..."
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Web service development and usage has shifted from simple information processing services to high-value business services that are crucial to productivity and success. In order to deal with an
increasing risk of unavailability or failure of mission-critical Web services we argue the need for advanced reservation of services in the form of derivatives. The contribution of this paper is
twofold: First we provide an abstract model of a market design that enables the trade of derivatives for mission-critical Web services. Our model satisfies requirements that result from service
characteristics such as intangibility and the impossibility to inventor services in order to meet fluctuating demand. It comprehends principles from models of incomplete markets such as the absence
of a tradeable underlying and consistent arbitragefree derivative pricing. Furthermore we provide an architecture for a Web service market that implements our model and describes the strategy space
and interaction of market participants in the trading process of service derivatives. We compare the underlying pricing processes to existing derivative models in energy exchanges, discuss eventual
shortcomings, and propose Wavelets as a preprocessing tool to analyze actual data and extract long- and short-term seasonalities.
, 1999
"... In this paper, we establish formulas for the configuration of a special class of irreducible representations of the Cuntz algebra ON, N = 2, 3,...,∞. These irreducible representations arise as
subrepresentations of naturally occurring representations of O_N acting in L2 (T) and arise from considerat ..."
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In this paper, we establish formulas for the configuration of a special class of irreducible representations of the Cuntz algebra ON, N = 2, 3,...,∞. These irreducible representations arise as
subrepresentations of naturally occurring representations of O_N acting in L2 (T) and arise from consideration of multiresolution wavelet filters.
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Sufficient conditions for Hausdorffness
up vote 0 down vote favorite
Let $(X,\tau)$ be a $T_1$ topological space and $Y\subset X$ a dense subspace which is completely metrizable. Are there any sufficient conditions to ensure that $(X,\tau)$ is Hausdorff using the
known facts?
EDIT: Here is an example of such a topological space which isn't Hausdorf, as Valerio asked.
Take $(Y,\|\cdot\|_1)$ and $(Y,\|\cdot\|_2)$ banach spaces such that there exists a sequence $(x_n)_{n\in\mathbb{N}}$ that converges to $x_1$ with respect with $\|\cdot\|_1$ and $x_2$ with respect to
$\|\cdot\|_2$ and $x_1\not=x_2$ (This norms exist). Then construct the normed vector space $(Y,\max\{\|\cdot\|_1,\|\cdot\|_2\})$; in this new normed vector space the sequence doesn't converge but is
Cauchy, so we complete it to the banach space $(X,\|\cdot\|_3)$. Now construct a new topology $\tau=\{A\ |\ A$ is open in $(X,\|\cdot\|_3) \land A\cap Y$ is open in $(Y,\|\cdot\|_1)\}$. Now $(X,\tau)
$ is an example of a topological space like the original problem. It is no Hausdorf because the sequence I define has two different limits, $x_1\in Y$ and $x_3\in X-Y$
I know it is a bit long and I didn't prove all the things I claimed to be true, but this is the exact space I was working on. I'm trying to check what pairs of norms in $Y$ produce that if a sequence
converges with respect to both of them, then it converges to the same element in $Y$.
topology gn.general-topology
do you have an example with a non Hausdorff space $X$? – Valerio Capraro Aug 8 '11 at 21:43
Isn't it easier to build an example by doubling a point of your favorite Polish space (e.g., the line with two origins)? – Clinton Conley Aug 8 '11 at 23:38
I used this example because is the space that appeared in my problem. I put it because it may be easier to answer the question for this particular space than the general problem. Like Paul Fabel
pointed out, the question isn't easy to answer in a general context. – dan232 Aug 8 '11 at 23:49
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1 Answer
active oldest votes
This answer begins with an easily understood fact and example followed by a more complicated example serving to illustrate why convenient answers to dan232's question can be
challenging to find.
FACT: If the 0-dimensional space X is T1, then X is T2.
Pf. Fix distinct points x and y. Since X is T1, X\y is open, and now obtain a clopen set U such that x is in U, and U is a subset of X\y. Note U and X\U are the desired open sets
which show X is T2.
Example 0: The indiscrete topology on a two point space shows a 0-dimensional space need not be T2.
Here is a more complicated example which answers Valerio's question, and shows a variety of nice properties can be inadequate to ensure that a T1 space is T2.
Example 1: X is a 1-dimensional space which enjoys the following properties:
Property 1: X is compact and dim(X)=1
up vote 2 down Property 2: Every compact subspace of X is a closed subspace of X (and in particular X is T1).
vote accepted
Property 3: There exists a point p in X such that Y=X\p is completely metrizable, (and in particular Y is open and dense in X).
Property 4: X is locally contractible.
Y is the union of countably many rays joined at a common point 0, and X is the one-point-compactification of Y (in the sense of Alexandroff).
To be precise, to obtain Y, consider the subspace of infinite rays emanating from 0 and passing through the standard `unit basis vectors' e1,e2,.... in the familiar Hilbert space l2
of square summable sequences of real numbers. Notice Y is a closed subspace of l2, and hence Y is completely metrizable. By definition Y will be an open dense subspace of its one
point (Alexandroff) compactification.
To obtain X, we create the special point at infinity p, and if U is a subset of Y union {p} such that p is in U, then U is open iff Y\U is a compact subspace of Y.
In particular, since Y is not locally compact at 0, X, the one point compactifcation of Y is not T2. However X enjoys the aformentioned listed properties.
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Building a polyhedron from areas of its faces
up vote 7 down vote favorite
Is there a known algorithm which, given a finite multiset (unordered list) of integers $A$, returns a yes/no answer for "Is there a polyhedron such that the multiset of areas of all its faces is
exactly $A$?"?
Is there a known general algorithm for $n$-dimensional polytopes?
combinatorial-geometry algorithms polytopes
There is one for two dimensions. If you don't mind prisms, you can likely adapt it to arbitrary dimensions. Gerhard "Ask Me About System Design" Paseman, 2011.11.30 – Gerhard Paseman Dec 1 '11 at
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1 Answer
active oldest votes
I can answer your question with the specialization to convex polyhedra and polytopes. Specializing further to $\mathbb{R}^3$, the result is that
$n \ge 4$ positive real numbers are the face areas of a convex polyhedron if and only if the largest number is not more than the sum of the others.
I wrote up a short note establishing this: "Convex Polyhedra Realizing Given Face Areas," arXiv:1101.0823. The result relies on Minkowski's 1911 theorem, which perhaps you know:
Theorem (Minkowski). Let $A_i$ be positive faces areas and $n_i$ distinct, noncoplanar unit face normals, $i=1,\ldots,n$. Then if $\sum_i A_i n_i = 0$, there is a closed convex
polyhedron whose faces areas uniquely realize those areas and normals.
up vote 13 This theorem reduces the problem to finding orientations $n_i$ so that vectors of length $A_i$ at those orientations sum to zero. And this is not difficult. Here is Figure 3 from my
down vote note from which you can almost infer the construction:
Minkowski's theorem generalizes to $\mathbb{R}^d$ and so does an analog of the above claim (but I did not work that out in detail in the arXiv note). In terms of an algorithm, the
decision question is linear in the number $n$ of facet areas, and even constructing the polyhedron is linear in $\mathbb{R}^3$, and likely $O(dn)$ in $\mathbb{R}^d$ (but again, I didn't
work that out).
But you don't mention the word "convex" in your post, so perhaps you are interested in nonconvex polyhedra and polytopal complexes?
Although I did not mention polygons, I suggested such in my comment above. There for n >=3 the condition is the same: no side must have length at least half the sum of all lengths.
For arbitrary dimensions, I suspect the same holds true of any nontrivial polytope, convex or not: no k-face has k-measure at least half the sum of the k-measures of all k-faces.
Triangular prisms should show that one cannot replace half by anything smaller. Gerhard "Ask Me About System Design" Paseman, 2011.11.30 – Gerhard Paseman Dec 1 '11 at 1:53
2 Let me mention that in a certain sense, the dual problem was resolved by Zil'berberg in 1962. This paper is not available in English I think, but is stated as Exercise 35.9 in my
book: math.ucla.edu/~pak/book.htm There, the areas are replaced by curvatures, which satisfy the Gauss-Bonnet formula and the proof is via an easy reduction to Alexandrov's "ray
theorem", which is dual to Minkowski's theorem (but neither easily implies another). – Igor Pak Dec 1 '11 at 2:16
(cont'd) One curious extension of Zil'berberg's proof is that the polytope can be made simple (for even number of vertices). I bet your theorem extends to have all polytopes
simpicial. – Igor Pak Dec 1 '11 at 2:17
I should be more careful in my statements. Let a polytope live in R^d. For positive integral k less than d there is a constant c_(k,d) such that the sum of the measures of all
k-faces of the polytope times that constant is greater than the k-measure of any single k-face. When k=d-1, I assert that the constant is 1/2. For smaller k, smaller constants are
possible, and are likely to be c_(k,d)=1/(1+d-k). Gerhard "Ask Me About System Design" Paseman, 2011.11.30 – Gerhard Paseman Dec 1 '11 at 2:17
As a remark to the case of an n dimensional non-convex polytope $P\subset \mathbb{R}^n$, Gerhard's condition is still necessary. Indeed, for any face f , the orthogonal projection
onto the hyperplane containing that face is 1-Lipschitz and maps the n−1 skeleton of P minus f surjectively onto f (the easy reason is: f has an inner and an outer side (essentially
by hypothesis on P, so that the line orthogonal to f at x∈f also meet ∂P in another y∉f). The analogous argument seems to work also for k-faces (giving the constant 1/2, possibly non
optimal) – Pietro Majer Dec 2 '11 at 11:55
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Sir Roger Penrose to deliver Pitcher Lecture
Renowned British mathematical physicist and Oxford Emeritus Professor Sir Roger Penrose will speak at 7:30 p.m. Wednesday in the Lewis Lab Auditorium as part of the
A. Everett Pitcher Lecture Series
His talk, titled "Before the Big Bang: An Outrageous Solution to a Profound Cosmological Puzzle,” is free and open to the public.
"We are very excited to have Professor Roger Penrose here at Lehigh to deliver this year's Pitcher Lectures,” says
Huai-Dong Cao
, professor and the A. Everett Pitcher Chair of Mathematics at Lehigh. “The public lecture on Wednesday should be a great opportunity for us to hear him elaborating on his new thinking about what may
have occurred prior to the so-called "big bang," a cosmic explosion that many scientists believe occurred billions of years ago, resulting in the creation of the universe."
In addition to Wednesday’s public lecture, Penrose will also speak on “Twistor Theory: Old Ideas and New" at 4:10 p.m. Thursday, March 16 in the Lewis Lab Auditorium, and at 2:10 p.m. Friday, March
17 in Neville Hall Auditorium 3.
Penrose is highly esteemed for his work in mathematical physics, in particular his contributions to general relativity and cosmology. His name is linked to several theories and discoveries such as
the Moore-Penrose inverse, the Penrose tiles, and Penrose diagrams. He also invented the well-known twistor theory, the mathematical theory that maps the geometric objects of the four-dimensional
space-time into the geometric objects in the 4-dimensional complex space with the metric signature.
Penrose has been awarded many prizes for his contributions to science, including the Eddington Medal of the Royal Astronomical Society and the prestigious Wolf Foundation Prize for Physics (both
along with Stephen Hawking, one of the world's leading theoretical physicists).
In 1994, he was knighted for his services to science. Penrose was also elected a Fellow of the Royal Society of London, Foreign Associate of the United States National Academy of Sciences, and was
awarded the De Morgan Medal for his wide and original contributions to mathematical physics. To quote the citation from the London Mathematical Society: "His deep work on General Relativity has been
a major factor in our understanding of black holes. His development of Twistor Theory has produced a beautiful and productive approach to the classical equations of mathematical physics. His tilings
of the plane underlie the newly discovered quasi-crystals."
Penrose recently received the 2006 Communications Award of the Joint Policy Board for Mathematics (JPBM) for the discovery of Penrose tilings, or shapes that tile the plane aperiodically, as well as
for the series of books that brought the subject of consciousness to the public in mathematical terms. In particular, the award citation praises his books
The Emperor's New Mind
The Road to Reality
. The citation states that the award is a tribute to the way Penrose has made the ideas behind high level mathematics accessible to large segments of the general public.
The series of lectures are held in honor of A. Everett Pitcher, who was secretary of the American Mathematical Society from 1967 until 1988. Pitcher served in the mathematics department at Lehigh
from 1938 until 1978, when he retired as Distinguished Professor of Mathematics. He will celebrate his 94th birthday in July.
The lecture is sponsored by the Pitcher Lecture Series of the Department of Mathematics, the Provost's Office, the College of Arts and Sciences, and the Department of Physics.
For additional information about Penrose's lectures, please call (610) 758-3731 or visit the
Web site
--Linda Harbrecht
Posted on Tuesday, March 14, 2006
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Homework Help
Posted by cl on Friday, May 8, 2009 at 1:11pm.
here is my source code for my program im trying to convert 10 units from furlong to meters the answer is supposed to be 2011.68 meters
i keep getting 2000. I know if i cant get this right the rest of the conversion will give the wrong answers anybody have any ideas?
Dim inch As Long
Dim fathom As Long
Dim foot As Long
Dim furlong As Long
Dim kilometer As Long
Dim meter As Long
Dim rod As Long
Dim miles As Long
Dim yard As Long
Dim Feet As Long
Dim Original As Long
Dim Desired As Long
Dim A(0 To 9) As Long
Dim B(0 To 9) As Long
Dim result As Long
Private Sub convertButton_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles convertButton.Click
Feet = Val(ConvertText.Text)
Original = Val(OUnitsTexT.Text)
Desired = Val(DUnitsText.Text)
inch = 0.0833 * Feet
yard = 3 * Feet
meter = 3.28155 * Feet
fathom = 6 * Feet
rod = 16.5 * Feet
furlong = 660 * Feet
kilometer = 3281.5 * Feet
miles = 5280 * Feet
A(1) = inch
A(2) = fathom
A(3) = foot
A(4) = furlong
A(5) = kilometer
A(6) = meter
A(7) = miles
A(8) = rod
A(9) = yard
B(1) = inch
B(2) = fathom
B(3) = Feet
B(4) = furlong
B(5) = kilometer
B(6) = meter
B(7) = miles
B(8) = rod
B(9) = yard
result = ((A(Original)) / ((B(Desired)) / Feet))
lengthText.Text = Val(result)
lengthText.Text = FormatNumber(result, 2, , , TriState.True)
• visual basic - SraJMcGin, Friday, May 8, 2009 at 2:41pm
I only know that taking this Conversion Chart, entering 6600 feet and changing to meters, I do get 2022.68.
This is definitely not my area so I can not help you set up a formula, if that is what you need.
• visual basic - RickP, Friday, May 8, 2009 at 7:04pm
The problem MIGHT be here.
result = ((A(Original)) / ((B(Desired)) / Feet))
Some languages (like C and C++) do integer division unless you explicitly make it do "normal" division. In other words, if you do 5/2 in C the result will be the integer 2, not the "floating
point number" 2.5.
• visual basic - RickP, Friday, May 8, 2009 at 7:08pm
I just figured it out: It's simpler than that.
You are declaring all of your numeric variables a LONG. LONG is an integer data type: it does not store anything to the right of a decimal point.
Try using DOUBLE (or better, DECIMAL, if your version supports it) for any numbers you need to have values to the right of the decimal point.
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Kurres exercises
Re: Kurres exercises
Actually, with a bit of afterthought, I'm perfectly happy with f(1) - 1 = f(1) + 1.
How about this? f(1) = √1
#4 doesn't hold up to that, unfortunately, unless you say:
f(x) = √1
For f(-x) < 0, x = √0.5
For f(-x) > 0, x = √z (such that 1/z = 0)
But that means making up crazy new Maths where we define the inputs for functions based on the output and 0z = 1.
And what kind of oddball would want to do that...?
Last edited by NullRoot (2007-11-22 11:07:57)
Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you
still can’t cope with is therefore your own problem.
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Limits and continuity...
June 23rd 2008, 03:48 PM #1
Limits and continuity...
I have got some HW problems based on limits and continuity...please help in solving those problems..
Q1) Find the limit of the following using L'Hospital rule....
a) lim x-> 0+ [sin(x) - x] / [e^x - 1]
b) lim x-> pi/2 [pi/2 - x] . tan(x)
Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1
I have got some HW problems based on limits and continuity...please help in solving those problems..
Q1) Find the limit of the following using L'Hospital rule....
a) lim x-> 0+ [sin(x) - x] / [e^x - 1]
b) lim x-> pi/2 [pi/2 - x] . tan(x)
Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1
Must you use L'hopital's?
Ok for the first one seeing that if we let $sin(x)-x=f(x)$ and $g(x)=e^x-1$
and $g(x)=f(x)=0$
Then $\lim_{x\to{0}}\frac{f(x)}{g(x)}=\lim_{x\to{0}}\fra c{f'(x)}{g'(x)}$
Edit: duplicated mathstuds fine work, sorry
L'Hopital's rule states that $\lim_{x\to x_0}\frac{f(x)}{g(x)} = \lim_{x\to x_0}\frac{f'(x)}{g'(x)}$ if $f(x_0)=0$ and $g(x_0) = 0$ or $f(x_0) = \pm \infty$ and $g(x_0) = \pm \infty$
Go for it. Question 1 is easy.
For question 2 start by taking the derivative of all parts of the inequality.
Last edited by badgerigar; June 23rd 2008 at 05:46 PM. Reason: fixed latex error, then saw previous post
I have got some HW problems based on limits and continuity...please help in solving those problems..
Q1) Find the limit of the following using L'Hospital rule....
a) lim x-> 0+ [sin(x) - x] / [e^x - 1]
b) lim x-> pi/2 [pi/2 - x] . tan(x)
Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1
The first one is of the form $\frac{0}{0}$ so now using L'hospitals rule we take a derivative of the numerator and the denominator to get
$\lim_{x \to 0^+}\frac{\cos(x)-1}{e^x}=\frac{\cos(0)-1}{e^{0}}=\frac{1-1}{1}=0$
For the 2nd rewrite it as $\frac{[x-\pi/2]\sin(x)}{\cos(x)}$
now as $x\to \frac{\pi}{2}$ this is of the form $\frac{0}{0}$
Using L.H again we get
$\lim_{x \to \frac{\pi}{2}}\frac{[x-\pi/2]\cos(x)+\sin(x)}{-\sin(x)}=\frac{[0]0+1}{-1}=-1$
For the 2nd question consider the function
$f(x)=\ln(x)$ f is continous on $[1,\infty)$ and differentiable on $(1,\infty)$ so it satisfies the hypothesis of the MVT
by the MVT $f(b)-f(a)=f'(c)(b-a)$ where $c \in (a,b)$
Let a=1 and b=x and we get
$\ln(x)-\ln(1)=\frac{1}{c}\left( x-1 \right)$ simplifying we get
$\ln(x) =\frac{x-1}{c} \iff c\ln(x)=x-1$
Since $c \in (1,x) \mbox{ or } 1< c < x$
$\ln(x) < x-1$
Also by the same reasoning $c < x \iff \frac{1}{x} < \frac{1}{c}$ so
$\frac{x-1}{x}< \frac{x-1}{c}=\ln(x)$
so finally we get
$\frac{x-1}{x} < \ln(x) < x-1$
Is it necessary L'Hôpital?
Q2 can also be killed as follows:
Since $\ln x=\int_{1}^{x}{\frac{dy}{y}},$ we have $\frac{1}{x}\le \frac{1}{y}\le 1\implies \frac{x-1}{x}\le \ln x\le x-1.\quad\blacksquare$ (This holds for $x>0.$)
Thanks a lot!
That really helps alot!
$\lim_{x \to \frac{\pi}{2}}\frac{[x-\frac{\pi}{2}]\sin(x)}{\cos(x)}=\frac{[\frac{\pi}{2}-\frac{\pi}{2}]\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2})}=\frac{[0]\cdot 1}{0}=\frac{0}{0}$
So now we can use L'hospitals rule.
Then we can take the derivatives to get what I did above.
I hope this helps.
Good luck.
Yes Sir! I got what you were trying to do after using the L'hospital Rule but I was asking that why are you manipulating the term? Since, in the problem it was given as ( pi/2 - x) sinx /
cosx...but you have mentioned that we can rewrite the function as
So why are you changing the sign here? And I guess that will change the answer. I got 1 from the function given in the original question and you are getting -1 from this one.
I am sorry to bother you. But I am just a little bit confused and want to learn if something I do not know.
Thanks so much once again!
The original problem was this
If we take the limit as $x \to \frac{\pi}{2}$
we get $0\cdot \pm \infty$
This doesnt quite fit the form to use L'hospitals rule.
We need to manipulate the equation into the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$
Hence I changed $[\frac{\pi}{2}-x]\tan(x)=\frac{[\frac{\pi}{2}-x]\sin(x)}{\cos(x)}$
Now this is of the form $\frac{0}{0}$
So from here we take the derivative of the numerator and the denomiator as per L.H's rule.
I guess I don't understand what you mean changing the sign?
The original problem was this
If we take the limit as $x \to \frac{\pi}{2}$
we get $0\cdot \pm \infty$
This doesnt quite fit the form to use L'hospitals rule.
We need to manipulate the equation into the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$
Hence I changed $[\frac{\pi}{2}-x]\tan(x)=\frac{[\frac{\pi}{2}-x]\sin(x)}{\cos(x)}$
Now this is of the form $\frac{0}{0}$
So from here we take the derivative of the numerator and the denomiator as per L.H's rule.
I guess I don't understand what you mean changing the sign?
You did
Which is the negative of the original problem.
Sorry, I didn't even notice that I switched them.
June 23rd 2008, 05:21 PM #2
June 23rd 2008, 05:24 PM #3
Senior Member
Dec 2007
June 23rd 2008, 05:35 PM #4
June 23rd 2008, 05:51 PM #5
June 23rd 2008, 06:02 PM #6
June 23rd 2008, 06:11 PM #7
June 23rd 2008, 06:23 PM #8
June 23rd 2008, 06:54 PM #9
June 23rd 2008, 07:16 PM #10
June 24th 2008, 03:52 AM #11
June 24th 2008, 10:26 AM #12
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August 13th 2005, 09:52 AM
Find the volume. Use 3.14 as pi
4.5ft h 2ft w
water storage tank is shaped like a circular cylinder with half of a sphere at each end. what is the volume of the tank if the cylindrical section is 100cm long with a 40cm diamerter. use 3.14
(pi) Round the answer to two decimal places
15 foot ladder leans against the barn, the bolltom of the ladder is 7ft from the building. how high is the top of the ladder. round to 2 decimal places.
please show me how to get these answers with the steps.
I can not get the hang of Geometry, I have never had the honor to have it in school
thanks in advance
August 20th 2005, 06:11 AM
Since we have TWO halves at either end,adding them will give us 1 complete sphere.
So now we have to find the volume of a sphere+volume of a cylinder.
Volume of a sphere is given by (4/3)x(pi)x(r^3)
Volume of a cylinder is given by (pi)x(r^2)x(h).
(Since diameter is given as 40cms and we know Diameter=2xRadius =>Radius r=20cms)
The ladder problem can be solved by using Pythagoras' Theorem which states that if a and b be the sides of a right angled triangle, the the third side(hypotenuse) is given by the root of sum of
squares of the other sides i.e. if C is the hypotenuse then c^2 = a^2 + b^2
You have to solve for (15)^2 = 7^2 + b^2
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Steenrod reduced power
From Encyclopedia of Mathematics
A stable cohomology operation Steenrod square, and which is a homomorphism
defined for every pair of topological spaces
2) if
3) if
4) (Cartan's formula)
5) (Adem's relation)
These properties are analogous to the corresponding properties of Steenrod squares, whereby the operation suspension and transgression.
The properties 1)–3) uniquely characterize
[1] N.E. Steenrod, D.B.A. Epstein, "Cohomology operations" , Princeton Univ. Press (1962)
[2] Matematika , 5 : 2 (1961) pp. 3–11; 11–30; 30–49; 50–102
For more references see Steenrod algebra.
How to Cite This Entry:
Steenrod reduced power. S.N. MalyginM.M. Postnikov (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Steenrod_reduced_power&oldid=15254
This text originally appeared in Encyclopedia of Mathematics - ISBN 1402006098
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[Numpy-discussion] find location of maximum values
questions anon questions.anon@gmail....
Mon Jan 9 22:40:02 CST 2012
Thank you, thank you, thank you!
I needed to find the max value (and corresponding TIME and LAT, LON) for
the entire month but I shouldn't have been using the tmax, instead I needed
to use the entire array. Below code works for those needing to do something
Thanks for all your help everyone!
maxtemp=TSFC.ravel()[maxindex] #or maxtemp=TSFC.max()
print maxindex, maxtemp
val=N.unravel_index(maxindex, TSFC.shape)
print listval
print latlocation, lonlocation
cdftime=utime('seconds since 1970-01-01 00:00:00')
print ncfiletime
On Tue, Jan 10, 2012 at 3:28 PM, Aronne Merrelli
> On Mon, Jan 9, 2012 at 7:59 PM, questions anon <questions.anon@gmail.com>wrote:
>> thank you, I seem to have made some progress (with lots of help)!!
>> I still seem to be having trouble with the time. Because it is hourly
>> data for a whole month I assume that is where my problem lies.
>> When I run the following code I alwayes receive the first timestamp of
>> the file. Not sure how to get around this:
>> tmax=TSFC.max(axis=0)
>> maxindex=tmax.argmax()
> You are computing max(axis=0) first. So, tmax is an array containing the
> maximum temperature at each lat/lon grid point, over the set of 721 months.
> It will be a [106, 193] array.
> So the argmax of tmax is an element in a shape [106,193] array (the number
> of latitude/number of longitude) not the original three dimension [721,
> 106, 193] array. Thus when you unravel it you can only get the first time
> value.
> I re-read your original post but I don't understand what number you need.
> Are you trying to get the single max value over the entire array? Or max
> value for each month? (a 721 element vector)? or something else?
> Cheers,
> Aronne
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion@scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
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RE: [SI-LIST] : Variability of Supply voltages
From: Ingraham, Andrew (Andrew.Ingraham@compaq.com)
Date: Tue Oct 10 2000 - 18:07:44 PDT
> Can we assume that when one power supply is at a maximum the other two
> supplies
> will be at a maximum?
No, as already described. There might be some interrelationships, but in
general, one should start by assuming that the three supply voltages are
unrelated; then later prove how and to what degree they aren't.
> Should we assume that all three supplies can be
> independent and we need to perform a maxtrix of 9 combinations to do
> a proper worst case analysis?
Well, that depends. If you know for some reason that the only interesting
cases to look at, are when all the supplies are at minimum, or all at
maximum, then maybe you could skip the other combinations, or some subset of
them. It may be that the other combinations are uninteresting as far as the
overall worst-case results are concerned. Or maybe not. A lot depends on
what goes on around the boundaries between the supply voltages. Margining
one low and the other high might cause significant pulsewidth shrinkage, for
example, which you would otherwise miss. To cover all the bases that
actually happen with your real circuit, you should do them all.
**** To unsubscribe from si-list or si-list-digest: send e-mail to
majordomo@silab.eng.sun.com. In the BODY of message put: UNSUBSCRIBE
si-list or UNSUBSCRIBE si-list-digest, for more help, put HELP.
si-list archives are accessible at http://www.qsl.net/wb6tpu
This archive was generated by hypermail 2b29 : Tue May 08 2001 - 14:29:42 PDT
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Discrete Math Tutors
Rowland Heights, CA 91748
Creative and Conceptual Math tutor
I have a bachelor of Science in Computer Science and Mathematics. I have been tutoring more than 4 years now. I tutor various math and science classes up to college level. My teaching idea is KTS,
meaning keep thing simple. That means, I try to break complex problems...
Offering 10+ subjects including discrete math
|
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Algorithms on seawater
This sections deals with algorithms for the calculation of properties of seawater. Most of the algorithms can also be found as fortran codes in the UNESCO technical paper 44 - 'Algorithms for
computation of fundamental properties of seawater' (published 1983). The here given algorithms are translated from the original fortran codes and are adopted to todays programming style guides.
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Area between curves
October 13th 2010, 08:56 PM #1
Sep 2010
Area between curves
Find the region between y=e^x and y=e^-x rotated around the y axis. The bound is x=1. Would I use shell method for this or washer?
Shells would work. I don't see how "washers" would be used here.
You could also do this is two separate parts. When x= 1, y= $y= e^{-x}= e^{-1}$ is a bound and at x=-1, $y= e^x= e^{-1}$ is a bound. That is, for y= 0 to $y= e^{-1}$ the right and left bounds of
the area are the straight lines x= 1 and x= -1. The figure is a cylinder, having area $\pi r^2 h= \pi(1)^2 e^{-1}= \pi e^{-1}$. You can do the volume from $y= e^{-1}$ up to y= 1 (where the two
graphs intersect) by discs.
October 13th 2010, 09:04 PM #2
Oct 2010
October 14th 2010, 04:38 AM #3
MHF Contributor
Apr 2005
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During the course of this century, gauge invariance has slowly emerged from being an incidental symmetry of electromagnetism to being a fundamental geometrical principle underlying the four known
fundamental physical interactions. The development has been in two stages. In the first stage (1916-1956) the geometrical significance of gauge-invariance gradually came to be appreciated and the
original abelian gauge-invariance of electromagnetism was generalized to non-abelian gauge invariance. In the second stage (1960-1975) it was found that, contrary to first appearances, the
non-abelian gauge-theories provided exactly the framework that was needed to describe the nuclear interactions (both weak and strong) and thus provided a universal framework for describing all known
fundamental interactions. In this work, Lochlainn O'Raifeartaigh describes the former phase.
O'Raifeartaigh first illustrates how gravitational theory and quantum mechanics played crucial roles in the reassessment of gauge theory as a geometric principle and as a framework for describing
both electromagnetism and gravitation. He then describes how the abelian electromagnetic gauge-theory was generalized to its present non-abelian form. The development is illustrated by including a
selection of relevant articles, many of them appearing here for the first time in English, notably by Weyl, Schrodinger, Klein, and London in the pre-war years, and by Pauli, Shaw, Yang-Mills, and
Utiyama after the war. The articles illustrate that the reassessment of gauge-theory, due in a large measure to Weyl, constituted a major philosophical as well as technical advance.
"The book thus performs a double service: it offers a rewarding description of the development of the gauge symmetry idea that is complete even without the original papers, and it makes those
original papers readily accessible to physicists and mathematicians. . . . This book represents an important contribution to the history of fundamental ideas in physics."--American Journal of Physics
Table of Contents
Subject Areas:
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Millbrae SAT Math Tutor
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CCSS.Math.Content.4.NF.A.1 - Wolfram Demonstrations Project
US Common Core State Standard Math 4.NF.A.1
Demonstrations 1 - 4 of 4
Description of Standard: Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even
though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
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How do I graph where this converges?
July 15th 2011, 02:16 PM #1
Aug 2010
How do I graph where this converges?
I need to sketch the graph to which this series converges:
$f(x) = \sum_{n=1}^{\infty} -\frac{4}{n\pi}(-1)^n sin(n\pi x)$
...and I don't have the foggiest idea of how to approach this. Can someone give me an idea of how to go about this?
Re: How do I graph where this converges?
Do you know what it converges to? This is a standard Fourier series, you can look it up in a table of FS.
How exactly is the question worded?
Last edited by CaptainBlack; July 16th 2011 at 07:32 AM.
Re: How do I graph where this converges?
After reading your post, I took a quick look, but did not come across any series that matched this one.
Regardless, unless I made a mistake, the only relevant part of the question is pretty much worded as I stated it in my original post: "Sketch the graph to which the Fourier series converges."
The original problem is this:
$f(x) = \begin{cases} x + 1, & -1 <= x < 0 \\ x - 1, & 0 <= x < 1 \end{cases} f(x + 2) = f(x)$ for all x
Re: How do I graph where this converges?
After reading your post, I took a quick look, but did not come across any series that matched this one.
Regardless, unless I made a mistake, the only relevant part of the question is pretty much worded as I stated it in my original post: "Sketch the graph to which the Fourier series converges."
The original problem is this:
$f(x) = \begin{cases} x + 1, & -1 <= x < 0 \\ x - 1, & 0 <= x < 1 \end{cases} f(x + 2) = f(x)$ for all x
Well you just have to sketch the curve given. Sketch the curve between -1 and 1, then as it is periodic with period 2 just tag copies of the sketch between -1 and 1 on at each end.
It is a saw tooth waveform going from $0$ at $x=-1$ to $1$ at $x=0_-$ and from $-1$ at $x=0$ to $0$ at $x=1$, then repeats ...
Note there appears to be a mistake somewhere, that series does not correspond to that function.
Last edited by CaptainBlack; July 16th 2011 at 07:34 AM.
Re: How do I graph where this converges?
I know how to sketch the curve of the initial set of equations. What I don't know is how to sketch the graph to which the resulting Fourier series converges.
I went over my work again, and I can't find any mistake. I'll show the major steps:
$p = 2$
$L = 1$
$a_0 = \frac{1}{L} \int_{-L}^L f(x) \; dx$
$a_0 = \frac{1}{1} \int_{-1}^1 x + 1 \; dx + \frac{1}{1} \int_{-1}^1 x - 1 \; dx$
$a_0 = \left[\frac{1}{2}x^2 + x + \frac{1}{2}x^2 - x \right]_{-1}^{1}$
$a_0 = 0$
$a_n = \frac{1}{L} \int_{-L}^L f(x) \; cos(\frac{n\pi x}{L}) \; dx$
$a_n = \int_{-1}^1 (x + 1) \; cos(n\pi x) \; dx + \int_{-1}^1 (x - 1) \; cos(n\pi x) \; dx$
$a_n = 2 \int_{-1}^1 x \; cos(n\pi x) \; dx$
$a_n = 2 \left[\frac{x}{n\pi} sin(n \pi x) + \frac{1}{n^2 \pi^2} cos(n \pi x)\right]_{-1}^1$
$a_n = 2[0] = 0$
$b_n = \frac{1}{L} \int_{-L}^L f(x) \; sin(\frac{n\pi x}{L}) \; dx$
$b_n = \int_{-1}^1 (x + 1) \; sin(n\pi x) \; dx + \int_{-1}^1 (x - 1) \; sin(n\pi x) \; dx$
$b_n = 2 \int_{-1}^1 x \; sin(n\pi x) \; dx$
$b_n = 2 \left[\frac{1}{n^2 \pi^2} sin(n \pi x) - \frac{x}{n \pi} cos(n \pi x)\right]_{-1}^1$
$b_n = 2 \left[- \frac{2}{n \pi} cos(n \pi) \right]$
$b_n = - \frac{4}{n \pi} (-1)^n$
$f(x) \sim \sum_{n = 1}^{\infty} \left[- \frac{4}{n \pi} (-1)^n sin(n \pi x) \right]$
Re: How do I graph where this converges?
I know how to sketch the curve of the initial set of equations. What I don't know is how to sketch the graph to which the resulting Fourier series converges.
I went over my work again, and I can't find any mistake. I'll show the major steps:
$p = 2$
$L = 1$
$a_0 = \frac{1}{L} \int_{-L}^L f(x) \; dx$
$a_0 = \frac{1}{1} \int_{-1}^1 x + 1 \; dx + \frac{1}{1} \int_{-1}^1 x - 1 \; dx$
$a_0 = \left[\frac{1}{2}x^2 + x + \frac{1}{2}x^2 - x \right]_{-1}^{1}$
$a_0 = 0$
$a_n = \frac{1}{L} \int_{-L}^L f(x) \; cos(\frac{n\pi x}{L}) \; dx$
$a_n = \int_{-1}^1 (x + 1) \; cos(n\pi x) \; dx + \int_{-1}^1 (x - 1) \; cos(n\pi x) \; dx$
$a_n = 2 \int_{-1}^1 x \; cos(n\pi x) \; dx$
$a_n = 2 \left[\frac{x}{n\pi} sin(n \pi x) + \frac{1}{n^2 \pi^2} cos(n \pi x)\right]_{-1}^1$
$a_n = 2[0] = 0$
$b_n = \frac{1}{L} \int_{-L}^L f(x) \; sin(\frac{n\pi x}{L}) \; dx$
$b_n = \int_{-1}^1 (x + 1) \; sin(n\pi x) \; dx + \int_{-1}^1 (x - 1) \; sin(n\pi x) \; dx$
$b_n = 2 \int_{-1}^1 x \; sin(n\pi x) \; dx$
$b_n = 2 \left[\frac{1}{n^2 \pi^2} sin(n \pi x) - \frac{x}{n \pi} cos(n \pi x)\right]_{-1}^1$
$b_n = 2 \left[- \frac{2}{n \pi} cos(n \pi) \right]$
$b_n = - \frac{4}{n \pi} (-1)^n$
$f(x) \sim \sum_{n = 1}^{\infty} \left[- \frac{4}{n \pi} (-1)^n sin(n \pi x) \right]$
Your limits of integration are wrong you should have sums of an integral from -1 to 0 and one from 0 to 1.
If you have done the working correct it will converge to the original function. If you are going to plot the partial sums you will need some computational system to do the calculations for any
thing other than two or three terms.
Re: How do I graph where this converges?
Sonofa... <smack>
Thank you for pointing that out! I completely missed that part of the technique when it was discussed.
Okay, so now that I've made that correction and redone everything, I now get this:
$f(x) \sim \sum_{n = 1}^\infty - \frac{2}{n \pi} \; sin(n \pi x)$
Looking at that, doesn't that just converge to zero?
Re: How do I graph where this converges?
Sonofa... <smack>
Thank you for pointing that out! I completely missed that part of the technique when it was discussed.
Okay, so now that I've made that correction and redone everything, I now get this:
$f(x) \sim \sum_{n = 1}^\infty - \frac{2}{n \pi} \; sin(n \pi x)$
Looking at that, doesn't that just converge to zero?
No it converges every where other than at x=0 to your function.
Re: How do I graph where this converges?
The sine function is limited to $[-1, 1]$, and $-\frac{2}{n \pi}$ approaches zero as $n$ approaches infinity. So why doesn't the series converge at zero?
p.s. Does the fourier series look correct to you, after my changes?
Re: How do I graph where this converges?
1. Why is it that:
$\sum_1^{\infty} \frac{1}{2^n} =2/3$
since your argument would imply it summed to zero (still does not sum to zero if we add alternating signs either).
2. Yes, it looks OK.
Re: How do I graph where this converges?
You can get WolframAlpha to plot the partial sums of this without difficulty:
sum for n=1 to 10 ( -2/(n*pi) sin(n*pi*x) ) - Wolfram|Alpha
Re: How do I graph where this converges?
Sorry, I realized I was processing the convergence the way I would a limit, rather than a series.
Thank you for your help!
Re: How do I graph where this converges?
Re: How do I graph where this converges?
It does converge to the midpoint of the jump, the thing is that it does not converge to the function value defined at the discontinuity (what I did say was probably was not clear enough on
July 15th 2011, 11:41 PM #2
Grand Panjandrum
Nov 2005
July 16th 2011, 06:03 AM #3
Aug 2010
July 16th 2011, 07:24 AM #4
Grand Panjandrum
Nov 2005
July 16th 2011, 09:20 AM #5
Aug 2010
July 16th 2011, 01:49 PM #6
Grand Panjandrum
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July 16th 2011, 02:40 PM #7
Aug 2010
July 16th 2011, 02:57 PM #8
Grand Panjandrum
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July 16th 2011, 03:06 PM #9
Aug 2010
July 16th 2011, 10:20 PM #10
Grand Panjandrum
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July 17th 2011, 02:25 AM #11
Grand Panjandrum
Nov 2005
July 17th 2011, 05:32 AM #12
Aug 2010
July 20th 2011, 08:31 PM #13
Aug 2010
July 20th 2011, 11:09 PM #14
Grand Panjandrum
Nov 2005
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"Attack" immune units and range
When figuring out what unit takes damage from an Attack, does the game just consider an unattackable unit as non-existent? (i.e. if only that unit is in row one and there is a unit in a back row,
does a range 1 attack do nothing or hit the second unit?)
I'm hoping it's the former, there's a lot of strategies that could be built around that. Even if it turns out otherwise, there's still the possibility to exploit other range-dependent skills.
Re: "Attack" immune units and range
Not exactly sure of what you are asking, since your sentence wasn't very clear. But some examples are below, hope it helps.
If your opponent have a monster in row 1 and a monster in row 3, and you have a monster in row 1 that only has a RNG of 1, your monster is only able to hit the opponent's monster in row 1.
If your opponent have a monster in row 2 and a monster in row 3, and you have a monster in row 1 that only has a RNG of 1, your monster is only able to hit the opponent's monster in row 2.
If your opponent have a monster in row 2 and a monster in row 3, and you have a monster in row 2 that only has a RNG of 1, your monster is only able to hit the opponent's monster in row 2.
If your opponent have a CLOSED monster in row 1 and a monster in row 3, and you have a monster in row 1 that only has a RNG of 1, your monster is able to hit the opponent's monster in row 3.
If your opponent have a monster in row 1 and a monster in row 3, and you have a monster in row 1 that only has a RNG of 2, your monster has a chance to hit the opponent's monster in row 1 or row 3.
(decided by the computer randomly)
Basically, RNG is the number of 'occupied' rows your monster can hit, with a RNG of 1, the monster can only hit the first row of monsters right in front of it, if it is a row of your own monsters,
then it's attack will be void.
Keyword is occupied, empty rows with closed or no monsters will not count towards RNG.
Logress wrote:Gota: I just looked at your lottery logs... I officially declare you the most unlucky man on earth.
Re: "Attack" immune units and range
Gota wrote:Not exactly sure of what you are asking, since your sentence wasn't very clear. But some examples are below, hope it helps
I think he was talking about the new falkow EX card : http://alteil.com/forum/index.php?f=12& ... =viewtopic
Alteil is pretty much well balanced. Sometimes you loose and sometimes the enemy wins.
Re: "Attack" immune units and range
slashzero wrote:When figuring out what unit takes damage from an Attack, does the game just consider an unattackable unit as non-existent? (i.e. if only that unit is in row one and there is a
unit in a back row, does a range 1 attack do nothing or hit the second unit?)
I'm hoping it's the former, there's a lot of strategies that could be built around that. Even if it turns out otherwise, there's still the possibility to exploit other range-dependent skills.
The card says "attack action" which means an action skill which will deal damage e.g. Brave Soldier's "slash"
Re: "Attack" immune units and range
I've used him before and discovered that it's basically the same as having a unit in the front row with infinite defense...unless your opponents use an ability, which are very common. Still, it's
useful for going against a deck that rely mainly on raw attack. Keep in mind that the units behind him can still be hurt by opponent's cards with enough range.
Re: "Attack" immune units and range
The short answer is if you have Ibert in your front row, and a bunch of units behind him, but no other units in the front row, then your enemy attacks with a RNG 1 unit in his front row, his attack
fails automatically. Basically, Ibert can't be a target but he still occupies the row, so everything behind him is out of range.
"Scissors are overpowered. Rock is fine." -Paper
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Brevet US5133023 - Means for resolving ambiguities in text based upon character context
This is a division of application Ser. No. 787,816 filed Oct. 15, 1985 now U.S. Pat. No. 4,754,489.
1. Field of the Invention
This invention relates in general to character recognition systems and relates more particularly to such systems having provision for resolving detected ambiguities in sensed characters.
2. Prior Art
A wide variety of pattern recognition systems are known in the art. Each such system optically receives data depicting a pattern to be recognized, and performs certain tasks on this pattern to
compare it to known patterns in order to "recognize" the input pattern. A basic flow chart depicting a pattern recognition system is shown in FIG. 1. The input pattern is the pattern which is to be
recognized. Digitizer 12 converts input pattern 11 to a series of bytes for storage in system memory 13. These bytes are typically binary in nature, reflecting the fact that input pattern 11 is
basically a black and white figure. Digitizers are well known in the art and typically are used in such devices as facsimile machines, electronic duplicating machines (as opposed to optical photocopy
machines) and optical character recognition systems of the prior art. Memory 13 can comprise any suitable memory device, including random access memories of well-known design. Segmentation 14 serves
to divide the image data stored in memory 13 into individual characters. Such segmentation is known in the prior art, and is described, for example, in "Digital Picture Processing," Second Edition,
Volume 2, Azriel Rosenfeld and Avinash C. Kak, Academic Press, 1982, specifically, Chapter 10 entitled "Segmentation".
Feature extraction 15 serves to transform each piece of data (i.e., each character) received from segmentation 14 into a standard predefined form for use by classification means 16, which in turn
identifies each character as one of a known set of characters.
Classification means 16 can be any one of a number of prior art identification means typically used in pattern recognition systems, including, more specifically, optical character recognition
systems. One such classification means suitable for use in accordance with the teachings of this invention is described in U.S. Pat. No. 4,259,661, issued Mar. 31, 1981 to Todd, entitled "Apparatus
and Method for Recognizing a Pattern". Identification means 16 is also described in "Syntactic Pattern Recognition and Applications," K. S. Fu, Prentice Hall, Inc., 1982, specifically, Section 1.6,
and Appendices A and B.
Postprocessing means 17 can be any one of a number of prior art postprocessing means typically used in pattern recognition systems, such as described in "n-Gram Statistics For Natural Language
Understanding And Text Processing" Suen, IEEE "Transactions on Pattern Analysis and Machine Intelligence," Vol. PAMI-1, No. 2, pp. 164-172, April 1979.
Output means 18 serves to provide data output (typically ASCII, or the like) to external circuitry (not shown).
Brief Description of the Drawings
FIG. 1 is a flow chart illustrating a typical prior art pattern recognition system;
FIG. 2a is an example of a prefix and a buffered block of possibility sets;
FIG. 2b is an example of a nonrecognition string;
FIG. 3 is a flow chart depicting the overall operation of the context module;
FIG. 4 is a flow chart depicting the operation of buffering a block of possibility sets;
FIG. 5 is a flow chart depicting the operation of choosing the winning string;
FIG. 6 is a flow chart depicting the operation of computing the confidence and status of each string candidate; and
FIG. 7 is a flow chart depicting the prefilter strings operation.
DESCRIPTION OF THE PREFERRED EMBODIMENT Preprocessing of Reference Data Computing Character String Probabilities
The context algorithm employed in the present invention, discussed later, assigns a probability value to each member of a set of character strings in order to determine which character string in the
set has the highest probability. As used in this specification, a character string is an ordered list of characters of some selected length. The probability value assigned to a character string is
computed by modeling the English language as a second order Markov process. That is ##EQU1## where x.sub.1 x.sub.2 x.sub.3 . . . x.sub.n is a string of n characters In the above equation, P(A/B)
denotes the relative probability of the event A given the event B. If, as above, the event A is a subset of the event B, then it is known that P(A/B)=P(A)/P(B). Using this fact, the above equation is
rewritten as ##EQU2## For any selected character string, after probability values for the appropriate character digrams and trigrams are obtained, the above equation is used to compute the
probability value of the selected character string. As used in this specification, a character digram is a string of two characters, and a character trigram is a string of three characters. The
problem of computing a probability value for a character string then reduces to the problem of computing probability values for the character digrams and character trigrams contained in the selected
character string.
Computing Trigram/Digram Probabilities
One standard method of obtaining the probability value of any character trigram (digram) is to use a precomputed character trigram (digram) probability table. Such a character trigram probability
table is constructed as follows. First, a large body of `typical` English text is obtained, which is referred to in this specification as the source tape. The character trigram probability table is
constructed so as to contain, for each possible character trigram, a value proportional to the number of occurrences of that selected character trigram in the source tape. The character digram
probability table is constructed in a similar way.
The method of this invention does not, however, use a character trigram table to obtain the probability of any character trigram, nor does it use a character digram table to obtain the probability
value of any character digram. Instead, it uses a total of eight different probability tables, in a manner fully described below, to obtain an estimate for the probability of any character digram or
trigram. This is done for two reasons: to minimize memory requirements and to improve statistical significance.
A character trigram probability table containing a probability value for any character trigram would require N.sup.3 *k bytes of memory, where N is the number of characters in the character set and k
is the number of bytes used to store a single probability value. If, for example, the character set contains 120 different characters and one byte is used to store each probability value, the
character trigram probability table would require approximately three megabytes of memory. In the method of this invention, the total amount of memory required to store the eight probability tables
is approximately 30K bytes.
As mentioned above, the second problem with using a character trigram (digram) probability table to obtain a probability value for any character trigram (digram) is that the probability values
generated by the source tape for low-probability trigrams (digrams) may not be statistically significant due to the limited size of the source tape. For example, referring to Table 1 below, the
number of occurrences in the source tape of each of three character trigrams is given.
TABLE 1______________________________________STRING # OCCURRENCES______________________________________535872______________________________________
The character trigram probability values for these three character trigrams, generated by the source tape, would indicate that `72 five times as probable as `58 The method of this invention assigns
more statistically significant probability values to low-frequency character digrams and trigrams.
As mentioned above, eight probability tables are computed. A formula is then derived to compute the probability of any character trigram or digram using values stored in these eight probability
tables. This formula is referred to in this specification as the `character ngram probability formula`. Prior to the construction of these tables, the following steps are taken:
Designate context independent groups
Designate generic letter independent nonletters
Define projections onto groups
Designate Context Independent Groups
Prior to the construction of probability tables, the first step is to designate certain groups of characters as `context-independent`. As used in this specification, a group of characters is
`context-independent` if the members of the group are assumed to be contextually indistinguishable. That is, a group of characters is context-independent if that group is assumed to satisfy the
context independent model, discussed below. In one embodiment, the groups of characters which are designated as context-independent are as given in Table 2 below.
TABLE 2______________________________________Group Name Characters in Group______________________________________<digit> group `0`, `1`, `2` . . . `9`<sentence-ending `!`, `?` punctuation> group<currency-suffix> ` group<arithmetic> group `+`, `=`, `______________________________________
The context independent model, used in the derivation of the character ngram probability formula, is the statement that
x and z are strings of characters
g is a context-independent group
y is a member of g
Stated informally, the context-independent model asserts that, given that a character is a member of a particular context-independent group, the probability that it is a particular member of that
group is independent of surrounding context.
The context-independent model is illustrated by examining the probability that a particular digit is the number `7` given that the digit is preceded by the letters `MB`:
The model is used to rewrite the above relative probability as follows: ##EQU3## The context-independent model asserts that the fact that a digit is preceded by the letters `MB` has no bearing on the
probability that the digit is the number `7`. If, for example, one out of every 10 digits in standard English usage is a `7`, that is if P(<7>/<digit>)=10%, then it is expected that
approximately 10% of all strings of the form <M><B><digit> end in the number `7`.
The designation of context independent groups is dependent on the given language. For example, in English, the character `.` is not included in the <sentence-ending punctuation> group
consisting of `!` and `?`. This is because the character `.` serves additional syntactic functions and so is sometimes contextually distinguisable from `!` and `?`. For example, the character `.` is
sometimes used as a decimal point and so is more likely to separate two digits than either `!` or `?`. In general, characters are grouped, when possible, into the same context-independent group in
order to reduce the amount of memory required to store the probability tables and in order to improve the statistical significance of the probabilities generated for low frequency character digrams
and trigrams.
Designate Generic Letter Independent Nonletters
Prior to the construction of probability tables, the second step is to designate certain nonletter characters as `generic letter independent`. As used in this specification, a selected nonletter
character is `generic letter independent` if it is assumed to satisfy the generic letter independent model, discussed below. In one embodiment, all nonletters except blank and apostrophe are
designated as generic letter independent. As used in this specification, the <generic letter independent> group is the set of all nonletters which have been designated as generic letter
The generic letter independent model, which is incorporated into the relative independent model discussed later, is the statement that
P(x.sub.1 yx.sub.2 /(x.sub.1 <g.l.i.>x.sub.2)=P(k.sub.1 yk.sub.2 /k.sub.1 >g.l.i.>k.sub.2)
x.sub.1 and x.sub.2 are strings of characters
k.sub.1 and k.sub.2 are obtained from x.sub.1 and x.sub.2 by replacing all occurrences of letters with the case of those letters
y is a nonletter that has been designated as generic letter independent
<g.l.i.> is the generic letter independent group
Stated informally, the generic letter independent model states that, given that a character belongs to the <generic letter independent> group, the probability that it is a particular member of
the group is dependent on only the case (i.e., uppercase or lowercase) of the surrounding letters.
This model is illustrated by examining the relative probability that a string of three letters is the string `MB?` given that the string consists of the letters `MB` followed by a character in the &
lt;generic letter independent> group:
In this example, the character `?` is assumed, as in one embodiment, to have been designated as generic letter independent. The model is used to rewrite the above probability as follows: ##EQU4## The
generic letter independent model asserts that the pair of capital letters `MB` is as likely to be followed by a question mark as is any other pair of capital letters. If, for example, 10% of all
character strings of the form <cap><cap><g.l.i.> are strings of the form <cap><cap><?>, then it is concluded that approximately 10% of all character strings of the
form <M><B><g.l.i.> are strings of the form <M><B><?>.
The designation of certain non-letter characters as generic letter independent is dependent on the given language. For example, in English, the character apostrophe is not designated as generic
letter independent. This is because the character apostrophe is more likely to separate certain pairs of letters (such as `n` and `t`) than other pairs of letters (such as `e` and `m`). In general,
nonletters are assumed to be generic letter independent whenever possible in order to reduce the amount of memory required to store the probability tables, discussed later, and to improve the
statistical significance of the probabilities generated for low frequency character trigrams.
Groups and Projections
The third and final step prior to the construction of probability tables is the creation of groups and projections. As used in this specification, a projection is a mapping from a set of characters
(domain of the projection) onto a set of character groups (range of the projection) with the following properties:
1. Each selected character in the domain of the projection is mapped onto a character group that contains the selected character.
2. No two groups in the range of the projection contain the same character.
Also as used in this specification, a character group containing a single character is referred to as an `identity` group.
A letter projection is a projection from the set of all letters onto a set of letter groups. In one embodiment, four letter projections are defined. These are:
letter-identity projection
letter-generic letter projection
letter-smallcap projection
letter-letter projection
The letter-identity projection maps each letter onto its corresponding identity letter group. The identity letter groups are ##EQU5## Thus, for example, the letter-identity projection of the
character lower-case `e` is the <e> group.
The letter-generic letter projection maps each letter onto its corresponding generic letter group. The generic letter groups are: ##EQU6## Thus, for example, the letter-generic letter projection of
both lowercase `e` and uppercase `E` is the <generic e> group.
The letter-smallcap projection maps all lower-case letters onto the <small> group and all upper-case letters onto the <cap> group.
Finally, the letter-letter projection maps all letters onto the single <letter> group.
A nonletter projection is a projection from the set of all nonletters onto a set of nonletter groups. In one embodiment, three nonletter projections are defined. These are:
nonletter-identity projection
nonletter-fine projection
nonletter-coarse projection
The nonletter-identity projection maps each nonletter onto its corresponding identity group. Thus, for example, the nonletter-identity projection of `?` is the <?> group.
The nonletter-fine projection maps each nonletter onto either a context-independent group of nonletter characters or an identity group. That is, if a selected nonletter is a member of a group of
nonletter characters which was designated, as described earlier, as a context-independent group, then that selected character is mapped onto that context-independent group. Otherwise, the selected
character is mapped onto an identity group containing that single character. In one embodiment the groups <digit> group, <sentence-ending punctuation> group, <currency-suffix>
group, and <arithmetic> group are, as earlier discussed, the groups designated as context-independent. Thus, in this embodiment, the nonletter-fine projection maps all digits onto the <digit
> group, the characters `!` and `?` onto the <sentence-ending punctuation> group, the characters ` characters `+`, `=`, ` nonletters onto identity groups.
The nonletter-coarse projection maps each nonletter which has been designated, as earlier discussed, as generic letter independent, onto the <generic letter independent> group and maps each
nonletter that has not been designated generic letter independent (i.e., has been designated as generic letter dependent) onto an identity group containing that selected character. As discussed
earlier, in one embodiment the characters blank and apostrophe are designated as generic letter dependent and all other nonletters are designated as generic letter independent. Thus, in this
embodiment, the nonletter-coarse projection maps the character blank onto the <blank> group, the character apostrophe onto the <apostrophe> group, and all other nonletters onto the <
generic letter independent> group.
The four letter projections and three non-letter projections are combined into nine different character projections. These are:
1. letter-identity/nonletter-coarse
2. letter-identity/nonletter-fine
3. letter-genericletter/nonletter-coarse
4. letter-genericletter/nonletter-fine
5. letter-smallcap/nonletter-coarse
6. letter-smallcap/nonletter-fine
7. letter-letter/nonletter-coarse
8. letter-letter/nonletter-fine
9. letter-identity/nonletter-identity
Thus, for example, the letter-smallcap/nonletter-coarse projection maps the letter lowercase `e` onto the <small> group and the nonletter ` ` onto the <generic letter independent> group.
Probability Tables
Eight probability tables are constructed using a source tape of `typical` English text. These are:
1. letter-letter/nonletter-coarse trigram table
2. letter-letter/nonletter-coarse digram table
3. letter-genericletter/nonletter-coarse trigram table
4. letter-genericletter/nonletter-coarse digram table
5. letter-smallcap/nonletter-fine trigram table
6. letter-smallcap/nonletter-fine digram table
7. letter-smallcap/nonletter-fine unigram table
8. individual character unigram table
The first seven tables are group probability tables. For example, the letter-genericletter/nonletter-coarse trigram table is a three dimensional probability table that contains a probability value
for each group trigram of letter-generic/nonletter-coarse groups, such as the group trigrams <generic n><apostrophe><generic t>, <generic t><generic h><generic e>,
and <generic t><generic letter independent><generic letter independent>. The last table in the list is a character probability table. For each selected character in the character
set, it gives the relative frequency of that character in the source tape.
The method used to construct the first seven probability tables is illustrated by discussing the method used to construct the letter-genericletter/nonletter-coarse trigram table. The other six tables
are constructed in a similar manner. First, the source tape of `typical` English text is converted into a list of letter-generic/nonletter-coarse groups. This conversion takes place by projecting
each character in the tape onto its corresponding group using the letter-genericletter/nonletter-coarse projection. Next, for each group trigram, its value in the trigram table is set to the number
of occurrences of that group trigram in the converted source tape. Finally, these frequency values are converted into probability (relative frequency) values by dividing each frequency value in the
table by the total number of group trigrams in the converted source tape. For example, ##EQU7##
Relative Independent Model
The derivation of the character ngram probability formula uses the projections defined above together with two models. The first model is the context independent model, discussed earlier. The second
model is the relative independent model. One embodiment of the relative independent model states that the letter-genericletter/nonletter-coarse projection and the letter-smallcap/nonletter-fine
projection are independent relative to the letter-letter/nonletter-coarse projection. As used in this specification, two projections A and B are independent relative to a third projection C if for
any string x of characters, ##EQU8## As used in this specification, if x is a string of characters and D is a projection, the notation D(x) denotes the string of groups obtained by projecting each
character in the string x onto a group using the projection D.
Thus, this embodiment of the relative independent model asserts that for any string x of characters, ##EQU9## where g-c is the letter-genericletter/nonletter-coarse projection
sc-f is the letter-smallcap/nonletter-fine projection
l-c is the letter-letter/nonletter-coarse projection
Stated informally, this relative independent model asserts that:
1. Given a string of characters containing letters, the projection of those letters onto case groups (<small> or <cap>) is independent of the projection of those letters onto generic
letter groups.
2. Given that a character is a member of the <generic letter independent> group, knowledge of what the surrounding generic letters are does not assist in computing the probability that the
character is a particular member of the <generic letter independent> group.
As is illustrated by an example below, this relative independent model is an extension of the generic letter independent model discussed earlier.
In the derivation of the character ngram probability formula, this relative independent model is used in conjunction with the fact that the letter-genericletter/nonletter-coarse projection and the
letter-smallcap/nonletter-fine projection `span` the letter-identity/nonletter-fine projection. As used in this specification, two projections A and B span a third projection C if any event expressed
in terms of the projection C can be reexpressed as the conjunction of events involving A and B. For example, if x is the string of characters `MB7`, its letter-identity/nonletter-fine projection is
the string of groups <M><B><digit>. The event that a string of characters is of the form <M><B><digit> can be reexpressed as the conjunction of the following two
1. The event that the character string is of the form <generic m><generic b><generic letter independent>. Note that this is the letter-genericletter/nonletter-coarse projection of
the character string `MB7`.
2. The event that the character string is of the form <cap><cap><digit>. Note that this is the letter-smallcap/nonletter fine projection of the character string `MB7`.
That is, the set of all character strings of the form <M><B><digit> is the intersection of the set of all character strings of the form <generic m><generic b><generic
letter independent> and the set of all character strings of the form <cap><cap><digit>. As used in this specification, the reexpression of an event expressed in terms of the
letter-identity/nonletter-fine projection as the conjunction of events expressed in terms of the letter-genericletter/nonletter-coarse projection and the letter-smallcap/nonletter-fine projection is
referred to as the `expansion` of the original event expressed in terms of the letter-identity/nonletter-fine projection.
Three examples are provided to illustrate the use of the relative independent model.
The first example applies the model to the problem of computing the relative probability that a given string of three characters is the string `The` given that it is a string of three letters:
The relative independent model is used to `factor` the above relative probability into the product of two relative probabilities as follows. ##EQU10## The justification for reexpressing (I.1) as
(I.2) is that the letter-genericletter projection and the letter-smallcap projection span the letter-identity projection. That is, the event
is expanded into the conjunction of events
<generic t><generic h><generic e> and <cap><small><small>
Thus, in going from (I.1) to (I.2), no assumptions about the statistics of the given language are used.
In going from (I.2) to (I.3), the relative independent model is used to approximate the probability (I.2) as the product of two probabilities. The meaning of this model can be better understood if it
is first noted that the probability (I.1) can also be expressed as follows: ##EQU11## Note that no assumptions about the statistics of the given language are made in (II). The only fact used is the
following fact about relative probabilities: If A and B are events with A a subset of B, then P(A/B)=P(A)/P(B). Comparing (I) and (II), it is seen that the relative independent model asserts that:
P(<T><h><e>/<generic t><generic h><generic e>)=P(<cap><small><small>/<letter><letter><letter>)
Thus, the relative independent model asserts that if, for example, 10% of all <letter><letter><letter> trigrams are <cap><small><small>, that is if:
then it is expected that approximately 10% of all <generic t><generic h><generic e> trigrams are cap-small-small, that is
P(<T><h><e>/<generic t><generic h><generic e>)=10%
The second example applies the relative independent model to the problem of computing the relative probability:
The computation proceeds as follows: ##EQU12##
In a manner similar to that described above in conjunction with the first example, the model asserts that if, for example, 10% of all <letter><apostrophe><letter> strings are <
cap><apostrophe><cap> strings, then it is expected that about 10% of all <generic n><apostrophe><generic> strings are of the form <cap><apostrophe><cap
In the English language, this relative independent model is most notably violated in the word:
Though in general, whether or not a character is uppercase is independent of what the generic letter is, this is not true for the string `I`. That is, though 20% of all <blank><letter>&
lt;blank> strings may be <blank><cap><blank> strings, it is not the case that approximately 20% of all <blank><generic i><blank> strings are <blank><I&
gt;<blank>. This case is discussed in more detail later.
The third example illustrates the fact that this relative independent model is an extension of the generic letter independent model discussed earlier. The relative independent model is applied to
That is, given a string of two letters followed by a nonletter that has been designated as a generic letter independent character, the model is used to compute the probability that the string
consists of the letters `MB` followed by a digit. This is done as follows. ##EQU13##
In going from (I.1) to (I.2) the event <M><B><digit> is expanded as the conjunction of events. In going from (I.2) to (I.3) the relative independent model is used. In a manner
similar to that described in conjunction with the first example above, the model is seen to assert that
P(<M><B><digit>/<generic m><generic b><g.l.i.>)=P(<cap><cap><digit>/<letter><letter><g.l.i.>)
That is, if for example 10% of all <letter><letter><generic letter independent> strings are <cap><cap><digit>, then 10% of all <generic m><generic b>&
lt;generic letter independent> strings are <cap><cap><digit>. The model asserts that whether a given generic letter independent character is, for example, a digit, is independent
of the surrounding generic letters. A digit is as likely to follow <generic m><generic b> as it is to follow any other generic letter pair.
Derivation of Character Ngram Probability Formula
The character ngram probability formula is derived for computing the probability of any character digram of trigram using the eight probability tables discussed earlier. The derivation assumes the
context independent model and the relative independent model. These models, and hence the derivation itself, can be appropriately modified for languages other than English. The formula derived can be
used to compute the probability of a string of characters of any length. For any integer n, the formula can be used to compute the probability of a string of characters of length n. Hence, for any
integer n, the formula can be used in conjunction with a context algorithm that models the English language as an nth order Markov process.
In the derivation, x is a string of characters of any length. The first step of the derivation expresses the probability of the character string in terms of relative probabilities, using group
projections. ##EQU14## where l-c(x) is the letter-letter/nonletter-coarse projection of x
i-f(x) is the letter-identity/nonletter-fine projection of x
i-i(x) is the letter-identity/nonletter-identity projection of x
The only fact used here is that if the event A is a subset of the event B, then P(A/B)=P(A)/P(B).
The second step is to rewrite (I.2) above using the fact that the letter-genericletter/nonletter-coarse projection and the letter-smallcap/nonletter-fine projection span the letter-identity/
nonletter-fine projection. Thus the letter-identity/nonletter-fine projection is expanded as the conjunction of projections as follows. ##EQU15## where g-c(x) is the letter-genericletter/
nonletter-coarse projection of x
sc-f(x) is the letter-smallcap/nonletter-fine projection of x
Substituting this identity for (I.2) in (I) above, the following identity is obtained ##EQU16##
The third step applies the relative-independent model in order to replace (II.2) above with the product of two relative probabilities (III.2) and (III.3), as follows ##EQU17##
The fourth step applies the context-independent model to (III.3). This model asserts that all multiple character nonletter-fine groups are context-independent, and hence ##EQU18## where the product
ranges over all characters x.sub.i in the string x which are members of context-independent groups. Substituting this identity for (III.4) in (III) above, the following identity is obtained. ##EQU19#
# where f(x.sub.i) is the nonletter-fine projection and the product ranges over those x.sub.i for which f(x.sub.i) is a context-independent group.
The fourth and final step is to replace all occurrences in (IV) of P(A/B) with P(A)/P(B) which is justified because in all cases, the event A is a subset of the event B. After making the substitution
and cancelling terms, the following equality is obtained. ##EQU20## where the x.sub.i in (V.3) and (V.4) range over all characters in the string x which are members of context-independent groups.
The above is the character ngram probability formula used to compute digram and trigram probabilities. A value for each term is determined by looking up the appropriate entry in the appropriate
probability table.
Three examples are provided illustrating the steps used in the previous derivation. These steps were:
1. express in terms of projections and relative probabilities
2. expand letter-identity/nonletter-fine projection
3. apply relative independent model
4. apply context independent model
5. express P(A/B) as P(A)/P(B)
EXAMPLE #1
The first example runs through the steps necessary to compute the probability of the string `The`:
Step 1
Express in terms of projections and relative probabilities. ##EQU21##
In this example, P(<T><h><e>/<T><h><e>) is equal to one, and hence is dropped from the equation.
Step 2
Expand <T><h><e>: ##EQU22##
Step 3
Apply relative independent model: ##EQU23##
Step 4
Apply context independent model:
This step is not applicable because there are no characters in the original string which belong to context independent groups.
Step 5
Replace all probabilities of the form P(A/B) with P(A)/P(B) and cancel terms: ##EQU24##
The first probability, P(<generic t><generic h><generic e>), is obtained from the letter-genericletter/nonletter-coarse trigram probability table. The second probability, P(<cap&
gt;<small><small>), is obtained from the letter-smallcap/nonletter-fine trigram probability table. The third probability, P(<letter><letter><letter>), is obtained from
the letter-letter/nonletter-coarse trigram probability table.
EXAMPLE #2
The second example computes the probability of the string `MB?`
It is assumed, for the sake of illustration, that the character `?` was designated as a member of the generic letter independent group of nonletters by the second model discussed earlier, and was
assigned to the context-independent group <sentence ending punctuation> group, consisting of the characters `?` and `!`, by the first model discussed earlier. The abbreviation <s.e.p.> is
used for the <sentence ending punctuation> group. The abbreviation <g.l.i.> is used for the <generic letter independent> group.
Step 1
Express in terms of projections and relative probabilities: ##EQU25##
Step 2
Expand <M><B><s.e.p.>: ##EQU26##
Step 3
Apply relative independent model: ##EQU27##
Step 4
Apply context independent model: ##EQU28##
Step 5
Rewrite P(A/B) as P(A)/P(B) and cancel terms: ##EQU29##
Each of the above probabilities is obtained from the appropriate probability table. The first probability is obtained from the letter-genericletter/nonletter-coarse trigram probability table. The
second probability is obtained from the letter-smallcap/nonletter-fine trigram probability table. The third probability is obtained from the letter-letter/nonletter-coarse trigram probability table.
The fourth probability is obtained from the character unigram probability table. The fifth probability is obtained from the letter-smallcap/nonletter-fine unigram table.
EXAMPLE 3
The third example computes the probability of the string `53
For the sake of illustration, it is assumed that ` to a context-independent group by the first model. It is also assumed that all three characters were assigned to the <generic letter independent&
gt; group of nonletters by the second model.
Step 1
Express in terms of projections and relative probabilities: ##EQU30##
Step 2
Expand <digit><digit>< ##EQU31##
Step 3
Apply relative independent model: ##EQU32##
Since (1) above is equal to one it can be dropped from the equation.
Step 4
Apply context independent model:
In this example, the digit group is assumed to be context independent. ##EQU33##
Step 5
Replace P(A/B) with P(A)/P(B) and cancel terms. ##EQU34##
Each of these probabilities is then obtained from the appropriate probability table.
If, in the above example, the character ` assigned to the context-independent group <currency suffix> group, consisting of ` the derivation would have given: ##EQU35##
Conversion to Confidences
Since multiplication and division are more costly than addition and subtraction, the probability tables are converted into confidence tables prior to the incorporation of these tables into the
context module. Each probability value in each probability table is replaced with a number proportional to the logarithm of that probability value. As used in this specification, the logarithm of a
probability value is referred to as a confidence value.
The character ngram probability formula is appropriately modified, using logarithms, so that instead of using probability values, it uses confidence values, as follows: ##EQU36## where the x.sub.i in
(1) and (2) range over all characters in the string x which are members of context-independent groups.
As used in this specification the above formula is referred to as the character ngram confidence formula.
Similarly, the formula used to compute the probability of any character string is appropriately modified, using logarithms, so that instead of using probability values of character trigrams and
digrams, it uses their confidence values, as follows: ##EQU37##
The input to the context module is either a single character or a set of more than one character candidates. If the input is a set of more than one character candidate then, as used in this
specification, each character candidate in the set is referred to as `ambiguous`. In one embodiment, associated with each ambiguous character candidate is an `a priori confidence value` which is set
by previous modules, such as the classification module, to a value proportional to the logarithm of the probability that the unknown input character is the character candidate, as that probability
was determined by such previous modules. In this embodiment, the a priori confidence values are incorporated into the computation of the probability of a character string as follows: ##EQU38## where
the sum in (1) ranges over those characters in the string which were ambiguous and conf (x.sub.i) is the a priori confidence of x.sub.i
FIG. 3 is a flowchart describing the overall operation of one embodiment of this invention. The operation begins with two initialization steps. The first initialization step stores the single
character `blank` in the prefix buffer. In general, the prefix buffer stores the last two characters output by the context module. However, if the last character output was the character `blank`, or
if no characters have yet been output, then the prefix stores the single character `blank`. The second initialization stage sets the last character type buffer to `miscellaneous`. The last character
type buffer stores information pertaining to the type of the last nonmiscellaneous character read. The character types are either letter, digit, or miscellaneous (all characters other than letters
and digits). In one embodiment, this buffer is used to assist in choosing between the word `I` and the word `1`.
Next, possibility sets are sequentially read. A possibility set contains one or more characters, together with an associated confidence value, which have been determined to possibly be the next
unknown character in the sequence of unknown characters being read. If the possibility set contains a single character which indicates unambiguously that the unknown character is that single
character contained in the possibility set, the prefix is updated by including the character contained in the possibility set just read as the most recently read character. The character type buffer
is then updated in accordance with the character type of the character contained in the possibility set. The possibility set is then output for use by other circuitry (not shown) and the next
possibility set is read.
On the other hand, if the possibility set read is not an unambiguous possibility set containing a single character, the context operation is performed in order to determine which of the characters
contained in the possibility set is most likely the unknown character being read and thus which should be the sole character remaining in the possibility set for output.
In this event, a block of possibility sets is buffered, as is more fully described in conjunction with FIG. 4. The buffered prefix, together with the buffered block of possibility sets, is used to
create a list of string candidates containing each permutation of the characters contained in the prefix and the buffered block of possibility sets. Table 3 below depicts the character strings that
are created from the prefix and block of possibility sets depicted in FIG. 2a.
TABLE 3__________________________________________________________________________Character GroupString StringCandidate Candidate Status__________________________________________________________________________ "Onlcns <blank><"><digit ><small><digit ><small><small><small> eliminated "OnlOns <blank><"><digit ><small><digit ><digit ><small><small> eliminated "OnlOns <blank><"><digit ><small><digit ><cap.sup. ><small><small> eliminated "Onlons <blank><"><digit ><small><digit ><small><small><small> eliminated "Onicns <blank><"><digit ><small><small><small><small><small> eliminated "OniOns <blank><"><digit ><small><small><digit ><small><small> eliminated "OniOns <blank><"><digit ><small><small><cap.sup. ><small><small> eliminated "Onions <blank><" ><digit ><small><small><small><small><small> eliminated "Onlcns <blank><"><digit ><small><small><small><small><small> eliminated "OnlOns <blank><"><digit ><small><small><digit ><small><small> eliminated "OnlOns <blank><"><digit ><small><small><cap.sup. ><small><small> eliminated "Onlons <blank><"><digit ><small><small><small><small><small> eliminated "Onlcns <blank><"><cap.sup. ><small><digit ><small><small><small> eliminated "OnlOns <blank><"><cap.sup. ><small><digit ><digit ><small><small> eliminated "OnlOns <blank><"><cap.sup. ><small><digit ><cap.sup. ><small><small> eliminated "Onlons <blank><"><cap.sup. ><small><digit ><small><small><small> eliminated "Onicns <blank><"><cap.sup. ><small><small><small><small><small> contender "OniOns <blank><"><cap.sup. ><small><small><digit ><small><small> eliminated "OniOns <blank><"><cap.sup. ><small><small><cap.sup. ><small><small> eliminated "Onions <blank><"><cap.sup. ><small><small><small><small><small> contender "Onlcns <blank><"><cap.sup. ><small><small><small><small><small> contender "OnlOns <blank><"><cap.sup. ><small><small><digit ><small><small> eliminated "OnlOns <blank><"><cap.sup. ><small><small><cap.sup. ><small><small> eliminated "Onlons <blank><"><cap.sup. ><small><small><small><small><small> contender "onlcns <blank><"><small><small><digit ><small><small><small> eliminated "onlOns <blank><"><small><small><digit ><digit ><small><small> eliminated "onlOns <blank><"><small><small><digit ><cap.sup. ><small><small> eliminated "onlons <blank><"><small><small><digit ><small><small><small> eliminated "onicns <blank><"><small><small><small><small><small><small> contender "oniOns <blank><"><small><small><small><digit ><small><small> eliminated "oniOns <blank><"><small><small><small> <cap.sup. ><small><small> eliminated "onions <blank><"><small><small><small><small><small><small> contender "onlcns <blank><"><small><small><small><small><small><small> contender "onlOns <blank><"><small><small><small><digit ><small><small> eliminated "onlOns <blank><"><small><small><small><cap.sup. ><small><small> eliminated "onlons <blank><"><small><small><small><small><small><small> contender__________________________________________________________________________
The block contains three ambiguous possibility sets. The first two ambiguous possibility sets contain three character candidates each. The third ambiguous poset contains 4 character candidates. Thus
there are 3 listed in the first column of Table 3. The second and third columns of Table 3 are discussed later in conjunction with the prescan operation of FIG. 7.
Next, the "winning" string of the string candidates contained in the list just prepared is determined, as is described in conjunction with FIG. 5. With the winning string selected, the prefix is
updated to include the two most recently read characters. However, if the most recently read character is a blank, the prefix is initialized to a single blank character. Also, the last character type
buffer is updated to correspond with the character type associated with the last nonmiscellaneous character read. Each ambiguous possibility set (i.e. a possibility set containing more than one
possible character) contained in the block of possibility sets is then modified to contain only that single character associated with that possibility set and the winning string candidate. The
possibility sets contained in the block of possibility sets are then output for use by additional circuitry (not shown). The operation is then reiterated by reading in the next possibility set,
determining if the possibility set is unambiguous, etc.
FIG. 4 is a flowchart depicting one embodiment of the step of buffering blocks of possibility sets, as described previously in conjunction with FIG. 3 (as shown in the example of FIG. 2a). As shown
in FIG. 4, a block, which consists of a plurality of possibility sets, is initialized to include the new, ambiguous possibility set just read. The next possibility set is then read and added to the
end of the block. In one embodiment of this invention, if the block ends in a blank character, and the last word (characters separated by a blank) contained in the block contains more than a single
character in width (such as the word "I"), the operation of buffering blocks of possibility sets is complete, and the return is made to the operation of FIG. 3. On the other hand, if the block does
not end in a blank character, or the last word in the block does not contain more than one character in width, but the last two possibility sets contained in the block are unambiguous (i.e., contain
only a single character each), the operation of buffering a block of possibility sets is complete. If not, a final test is made to determine if the size of the block of buffered possibility sets has
reached preset limit (in one embodiment 15 possibility sets), in which case the operation of buffering a block of possibility sets is complete. If not, the operation of buffering a block of
possibility sets continues with the step of reading in the next possibility set.
In this manner the operation, such as is shown in FIG. 4, stores a plurality of possibility sets in a buffered block for further analysis, commencing with the step of building a list of string
candidates (FIG. 3) containing each permutation of the characters contained in the buffered possibility sets.
FIG. 5 is a flowchart depicting the step of choosing a winning candidate string, as previously described in conjunction with FIG. 3. First, as is described more fully below in conjunction with the
flow chart of FIG. 7, the strings are prefiltered, if desired, for example, in order to eliminate from the list of string candidates certain strings which are highly unlikely.
Following the prefilter operation, a check is made to determine the number of candidate strings left. In the example depicted in Table 3, shown earlier, 28 of the original 36 string candidates have
been eliminated by the prefilter operation. These strings are deleted from the list of string candidates. Table 4 depicts the 8 string candidates that remain after the prefilter operation.
TABLE 4______________________________________String Candidate Status Confidence______________________________________ "Onicns eliminated "Onions contender 70 "Onlcns eliminated "Onlons contender 50 "onicns eliminated "onions contender 65 "onlcns eliminated "onlons contender 45______________________________________
If exactly one candidate string is left after the prefilter operation, this candidate string is selected as the appropriate string and a return is made to the context algorithm of FIG. 3, which then
updates the possibility sets in accordance with the information contained in the single remaining candidate string.
The prefilter operation always leaves at least one string; in one embodiment of this invention, if the prefilter operation would normally delete all candidate strings, the prefilter operation has not
assisted in identifying the unknown text being read and thus all string candidates remain.
If more than one string candidate remains, the context mode flag is set to either full language context or structural context. In one embodiment of this invention, the setting of this flag is
determined by an operator-actuated switch available on the front panel of the machine. If the context mode flag is set to "full language," then the confidence value assigned to a character digram or
trigram is computed using the character ngram confidence formula earlier described. If, on the other hand, the context mode is set to "structural" then the confidence value is computed using the
`structural character ngram formula` which, in one embodiment, is
Log P(x)=Log P(sc-f(x))-sum {Log (size(sc-f(x.sub.i)))}
x is a string of candidates
sc-f(x) is the letter-smallcap/nonletter-fine projection of x
x.sub.i is a character in the character string x
size (sc-f(x.sub.i)) is the number of characters in the letter-smallcap/nonletter-fine projection group of x.sub.i
The character ngram confidence formula uses knowledge of probability statistics in a given selected language. If the input document is written in a language whose probability statistics differ
significantly from the probability statistics of the selected language, then the context mode is set to `structural`. In this mode, the structural character ngram formula is used. Thus, in this mode,
knowledge of generic letter digram and trigram probabilities is not used, and the members of any letter-smallcap/nonletter-fine projection group are assumed to be uniformly distributed.
Next, the status (i.e., contender or eliminated) and the confidence value of each string candidate is computed, as is more fully described in conjunction with FIG. 6.
In one embodiment of this invention, a separate flag is available to set the recognition mode as being either "recognize only if certain", or "always take best guess". The setting of this flag is
determined by an operator-activated switch available on the front panel of the machine. If this flag is set to "recognize only if certain" and if the context module decides there is some uncertainty
as to which character candidate in the possibility set is the correct choice, then the context module outputs a possibility set containing a default, nonrecognition character, such as "@".
Referring again to FIG. 5, if the recognition mode is set to "recognize only if certain," then the number of candidate strings whose status is "contender" is determined. If the number of "contender"
strings is equal to one, that is, if all but one candidate string has been eliminated by either the prefilter operation or the compute status and confidence operation, then the remaining string
candidate, having its status equal to "contender," is selected as the winning string candidate, and a return is made to the operation of FIG. 3. Conversely, if the recognition mode is equal to the
"recognize-only-if-certain" mode, and it is not the case that all but one candidate string has been eliminated, a "nonrecognition string" is created as the winning string. This nonrecognition string
is constructed as follows. For each unambiguous possibility set forming the buffer block, the character in the corresponding slot of the nonrecognition string is set equal to the character contained
in its corresponding possibility set. The characters in the nonrecognition string corresponding to ambiguous possibility sets (containing more than a single possible character) are set to a default
nonrecognition ymbol, such as the "@". Referring to Table 4, more than one string candidate has the status value "contender." If the recognition mode is set to the "recognize-only-if-certain" mode,
then the nonrecognition string is created as is depicted in FIG. 2b.
On the other hand, if the recognition mode is not set equal to the "recognize-only-if-certain" mode, i.e., the recognition mode is set equal to the "best guess" mode, a decision is made whether all
string candidates have been "eliminated". If not, the string candidate whose status is still "contender", and has the highest confidence value, is selected as the winning string candidate, and a
return is made to the context operation of FIG. 3. Referring to Table 4, there are 4 candidate strings whose status is "contender." Of these, the string
<blank>"Onions is the string with the highest confidence and so is chosen as the winning string. Conversely, if all string candidates have a status of "eliminated", and the context mode is
"structured context" rather than "full-language context", the winning string is selected such that each ambiguous character is set equal to the first character contained in the associated possibility
set, which corresponds to that character in the associated possibility set having the highest confidence value assigned during the character recognition operation. Conversely, if the context mode is
"full-language", the context mode is reset to the "structural-context" mode and another iteration of the choose winning string candidate operation of FIG. 5 is performed, beginning with the step of
computing the status and confidence of each string candidate. This reiteration after changing the context mode flag to "structural context" may provide a different result because the confidence of
each string candidate is dependent upon the context mode, as has been previously described. In this manner, the context module is able to process input documents containing words that do not conform
to the statistics of the selected language. For example, if the input document contains the name
and, for each `z` in the above work, the input to the context module is a possibility set containing the characters `z`, `Z`, and `?`, then all strings will be assigned a status of `eliminated` by
the first iteration that uses a context mode of `full language`, but the correct string will be selected by the second iteration that uses a context mode of `structural context`.
Compute Confidence
The compute confidence of string candidate operation of FIG. 5 is now described in more detail in conjunction with the flow chart of FIG. 6. The compute confidence of string candidate operation is
called as a subroutine from the choose winning string operation of FIG. 5 and serves to establish a confidence value for each string candidate which has previously been established during the build
list of string candidates operation of FIG. 3 and has not been deleted by the prefilter operation of FIG. 7. The first step is to select the first such candidate string. Next, in one embodiment of
this invention in which an extended ASCII code is used which includes certain specialized characters such as "ff" and "ffi", these characters are broken down into two and three characters,
respectively, in order that each such character may be treated individually. Next, the string confidence is set to zero, and then the string status is set to "contender". A window is set to
correspond to the first three characters of the string. Naturally, as will be appreciated by those of ordinary skill in the art, a window other than length 3 can be used, although for windows of
greater length, a correspondingly greater amount of memory is required in order to store the corresponding probability tables. Next, the trigram confidence value of the three characters contained
within the window is computed according the the equations previously given. That is, if the context mode is set equal to the "full language" context mode, then the character ngram confidence formula
is used. If the context mode is set equal to the "structural" context mode, then the structural character ngram confidence formula is used. Next, if it is determined that the trigram probability is
zero (i.e., the trigram is impossible), the status of the string being analyzed is set to "eliminated" and the operation continues if there are further strings to be obtained. If there are no further
strings to be obtained, a return is made to the choose winning string algorithm of FIG. 5.
On the other hand, if the trigram probability is not equal to zero, indicating that this trigram is possible, the trigram confidence is added to the string confidence value. As described in this
specification, confidence values during this operation are added, since they are stored in probability tables as logarithms of probabilities. Next, it is determined if the window currently contains
the last three characters of the string candidate being analyzed. If so, the operation reiterates if there are further strings to be analyzed or makes an exit to the choose winning string operation
of FIG. 5 if there are no further string candidates to be analyzed.
On the other hand, if the window does not contain the last three characters of the string, the window is shifted to the right by one character, and the trigram confidence of the characters contained
in the window is again computed in accordance with the equations previously described. If the trigram confidence is determined to be impossible, the status of the string candidate being examined is
set to "eliminated" and the operation continues if there are any remaining strings to be analyzed. On the other hand, if the trigram confidence is not impossible, the digram confidence of the first
two characters in the window is calculated as previously described, and the value equal to the trigram confidence minus the digram confidence is added to the string candidate confidence value in
order to provide the logarithm of the relative probability of the characters in the current window given the characters in the previous window, as was earlier described. Next, the operation branches
back to the step of determining if the window contains the last three characters of the string, and if not, further analysis is made after shifting the window to the right by one character.
FIG. 7 depicts the prefilter operation discussed earlier in conjunction with the flow chart of FIG. 5.
The first step is the coarse structural filter. The coarse structured filter uses a precalculated coarse structural trigram table, which in one embodiment is constructed as follows. The table has an
entry of either `good` or `bad` for each trigram of letter-smallcap/nonletter-fine projection groups. The value of a selected group trigram is set to `good` if the corresponding entry in the
letter-smallcap/nonletter-fine trigram probability table indicates that the selected group trigram has a non-zero probability, otherwise the value is set to `bad`. After this, certain group trigrams
are deemed `structurally improbable` and their values are reset from `good` to `bad`. In one embodiment, the following group trigrams are deemed structurally improbable.
TABLE 5______________________________________ <small><small><cap> <small><cap.sup. ><small> <small><cap.sup. ><cap> <cap.sup. ><small><cap> <cap.sup. ><cap.sup. ><small> <small><small><digit > <small><cap.sup. ><digit > <small><digit ><small> <small><digit ><cap> <small><digit ><digit > <cap.sup. ><small><digit > <cap.sup. ><cap.sup. ><digit > <cap.sup. ><digit ><small> <cap.sup. ><digit ><cap> <cap.sup. ><digit ><digit > <digit ><small><small> <digit ><small><cap> <digit ><small><digit > <digit ><cap.sup. ><small> <digit ><cap.sup. ><cap> <digit ><cap.sup. ><digit > <digit ><digit ><small> <digit ><digit ><cap> <blank><small><cap> <blank><small><digit > <blank><cap.sup. ><digit > <blank><digit ><small> <blank><digit ><cap> <small><cap.sup. ><blank> <small><digit ><blank> <cap.sup. ><digit ><blank> <digit ><small><blank> <digit ><cap.sup. ><blank>______________________________________
The coarse structural filter operation uses the coarse structural trigram table to assign a temporary status value of either `contender` or `eliminated` to each candidate string. This is done as
follows. For each selected character candidate string, a corresponding group candidate string is created, using, in one embodiment, the letter-smallcap/nonletter-fine projection. The group candidate
string is then examined. If it contains any group trigram whose value, in the coarse structural trigram table, is `bad`, then the temporary status associated with the selected character candidate
string is set to "ELIMINATED", otherwise it is set to "CANDIDATE". Referring to Table 3, the temporary status assigned to the string <blank>Onions is set to "eliminated" because the value of
the group trigram <digit><small><digit> in the coarse structural trigram table is `bad`. Of the 36 original character strings in the example depicted in Table 3, 28 are assigned a
temporary status value of "eliminated" by the coarse structural filter. After the coarse structural filter operation assigns a temporary status value to each character candidate string, if it is not
the case that all strings have been assigned a temporary status value of "eliminated", then those strings whose status value is "eliminated" are deleted. Table 4 depicts those candidate strings which
were not deleted by the coarse structural filter operation.
In one embodiment, after the coarse structural filter operation is performed, the next step is the I/i filter operation. In one embodiment, the word `i` is assigned a higher probability than the word
`I` by the character ngram confidence formula. To prevent the wrong string from being chosen, the I/i filter operation checks to see if there are candidate strings containing `i` and other candidate
strings containing `I` in the corresponding locations. If so, it deletes those candidate strings containing `i`.
In one embodiment, after the I/i filter operation is performed, the I/1 filter operation is performed. In one embodiment, the word `I` is assigned a higher probability value than the word `1`. The I/
1 filter operation checks to see if there are candidate strings containing `I` and other candidate strings containing `1` in the corresponding locations. If so, the I/1 filter operation uses
information contained in the prefix, the buffered block of posets, and the last character type buffer to possibly delete all of the candidate strings containing `I`. In one embodiment, if the last
character type is `digit`, or the prefix contains a digit or there is an unambiguous possibility set in the buffered block of possibility sets that contains a digit, then the candidate strings
containing `I` are deleted.
While this specification illustrates specific embodiments of this invention, it is not to be interpreted as limiting the scope of the invention. Many embodiments of this invention will become evident
to those of ordinary skill in the art in light of the teachings of this specification. For example, although the method described earlier used projections, models and group probability tables to
derive a formula that is used to assign a probability to any character ngram, the present invention can be appropriately modified to assign a probability value to any object ngram where strings of
such objects obey statistical rules. For example, the method of this invention can be used in a speech recognition system where an object is a phoneme and certain combinations of phonemes are more
probable than other combinations.
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Abstract for rosti_thesis
Department of Engineering
Abstract for rosti_thesis
PhD thesis, University of Cambridge, 2004
A-V.I. Rosti
May, 2004
Currently the most popular acoustic model for speech recognition is the hidden Markov model (HMM). However, HMMs are based on a series of assumptions, some of which are known to be poor. In
particular, the assumption that successive speech frames are conditionally independent given the discrete state that generated them is not a good assumption for speech recognition. State space models
may be used to address some shortcomings of this assumption. State space models are based on a continuous state vector evolving through time according to a state evolution process. The observations
are then generated by an observation process, which maps the current continuous state vector onto the observation space. In this work, the state evolution and observation processes are assumed to be
linear and noise sources are distributed according to Gaussians or Gaussian mixture models. Two forms of state evolution processes are considered.
First, the state evolution process is assumed to be piece-wise constant. All the variations of the state vector about these constant values are modelled as noise. Using this approximation, a new
acoustic model called the factor analysed HMM (FAHMM) is presented. In the FAHMM a discrete Markov random variable chooses the continuous state and the observation process parameters. The FAHMM
generalises a number of standard covariance models such as the independent factor analysis, shared factor analysis and semi-tied covariance matrix HMMs. Efficient training and recognition algorithms
for the FAHMMs are presented along with speech recognition results using various configurations.
Second, the state evolution process is assumed to be a linear first-order Gauss-Markov random process. Using Gaussian distributed noise sources and a factor analysis observation process this model
corresponds to a linear dynamical system (LDS). For acoustic modelling a discrete Markov random variable is required to choose the LDS parameters. This hybrid model is called the switching linear
dynamical system (SLDS). The SLDS is related to the stochastic segment model, which assumes that the segments are independent. In contrast, for the SLDS the continuous state vector is propagated over
the segment boundaries, thus providing a better model for co-articulation. Unfortunately, exact inference for the SLDS is intractable due to the exponential growth of posterior components in time. In
this work, approximate methods based on both deterministic and stochastic algorithms are described. An efficient proposal mechanism for Gibbs sampling is introduced along with application to
parameter optimisation and N-best rescoring. The results of medium vocabulary speech recognition experiments are presented.
Keywords: Speech recognition, acoustic modelling, hidden Markov models, state space models, linear dynamical systems, expectation maximisation, Markov chain Monte Carlo methods
| (ftp:) rosti_thesis.pdf | (http:) rosti_thesis.pdf | (ftp:) rosti_thesis.ps.gz | (http:) rosti_thesis.ps.gz |
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BJT single stage amplifier we write gm=Ic/VT
Each device has g[m] value equal to its own I[c] divided by V[T], where V[T] is 25.7 mV for each device. For a Darlington, if the input device is no. 1, and output device is no. 2, then I[e1] = I
[b2], I[c1] = α[1]*I[e1], I[c2] = β[2]*I[b2], or I[c2] = α[2]*I[e2], .
This should get you started.
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[SciPy-user] noncentral F distribution?
Neal Becker ndbecker2@gmail....
Wed May 27 18:38:29 CDT 2009
josef.pktd@gmail.com wrote:
> On Wed, May 27, 2009 at 7:16 PM, Neal Becker <ndbecker2@gmail.com> wrote:
>> Does scipy have non central F distribution? (I need cdf for that)
>>>> print scipy.stats.ncf.extradoc
> Non-central F distribution
> ncf.pdf(x,df1,df2,nc) = exp(nc/2 + nc*df1*x/(2*(df1*x+df2)))
> * df1**(df1/2) * df2**(df2/2) * x**(df1/2-1)
> * (df2+df1*x)**(-(df1+df2)/2)
> * gamma(df1/2)*gamma(1+df2/2)
> * L^{v1/2-1}^{v2/2}(-nc*v1*x/(2*(v1*x+v2)))
> / (B(v1/2, v2/2) * gamma((v1+v2)/2))
> for df1, df2, nc > 0.
>>>> scipy.stats.ncf.cdf
> <bound method ncf_gen.cdf of <scipy.stats.distributions.ncf_gen object
> at 0x021DDE90>>
> note 3rd or 4th moments are wrong
> Josef
I found the page:
but I don't know what the parameters mean.
I was looking for something like:
There, a ncf is constructed with 3 parameters, v1, v2, lambda.
Then the cdf is given as a function of a single variable, x.
In scipy.stats.ncf, there are many constructor parameters. Which correspond
to the v1,v2,lambda I was looking for?
scipy.stats.ncf(momtype=1, a=None, b=None, xa=-10.0, xb=10.0, xtol=1e-14,
badvalue=None, name=None, longname=None, shapes=None, extradoc=None)
In scipy.stats.ncf the cdf has
again, I don't know what they mean. I think x is my x, but I don't know
what the others are.
I haven't used scipy stats before, so maybe I'm just not familiar with the
interface. (I'm hoping I don't have to go back to program in c++ for this
More information about the SciPy-user mailing list
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Which of the following is not an equivalent form of the compound inequality x + 6 < 9 or x + 6 >= 15 A) x < 3 or x >= 9 B) A number line with an open circle on 3, a closed circle on 9, and shading in
between. C) x >= 9 or x < 3 D) A number line with an open circle on 3, shading to the left, and a closed circle on 9. shading to the right.
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its B
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Choose the correct description of the graph of the compound inequality 2x + 5 > 9 and 5x <= 30 A number line with an open circle on 2, shading to the left, and a closed circle on 6, shading to
the right. A number line with an open circle on 2, a closed circle on 6, and shading in between. A number line with a closed circle on 2, shading to the left, and an open circle on 6, shading to
the right. A number line with a closed circle on 2, an open circle on 6, and shading in between.
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diagram for previous 1 :D
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answer is B for 2nd 1
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A number line with an open circle on 2, a closed circle on 6, and shading in between.
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Which of the following is not an equivalent form of the compound inequality x + 15 <= 12 and x + 15 >= 5 12 >= x + 15 >= 5 −3 >= x >= −10 A number line with a closed circle on −10, a closed
circle on −3, and shading in between. A number line with a closed circle on −10, shading to the left, and a closed circle on −3, shading to the right.
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the one right there^^ not the original lol
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wait let me solve
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okie dokies
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A number line with a closed circle on −10, shading to the left, and a closed circle on −3, shading to the right. this is false
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okay m off for tonight :) good night
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okie dokies! night! thank you!!
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A Linear Algebra Problem
up vote 7 down vote favorite
Given a matrix $A\in \mathbb{R}^{n\times n}$, I am looking for a symmetric matrix $S\in\mathbb{R}^{n\times n}$ such that
$$ S A + A^T S = I $$
$A$ can be assumed to be regular (with positive determinant, if this is of any help).
The difficulty is of course that $S$ must be symmetric, otherwise one could simply take $2S = A^{-T}$. In principle this is a linear equation with $\frac{n(n+1)}{2}$ unknowns and this can be solved
for $S$.
Is there a nicer way to find $S$ such as a closed solution formula using some factorization? Has this problem been studied anywhere?
matrices linear-algebra
Do you already know (and is it true at all) that the system of linear equations is uniquely solvable? – darij grinberg Jan 6 '12 at 14:27
Yes, this follows from geometric considerations (which are a bit too lengthy for me to reproduce here). – Philipp Jan 6 '12 at 14:39
2 Why is there a $S^T$ in the equation if $S$ is symmetric? – Federico Poloni Jan 6 '12 at 14:59
fair enough; I have deleted the transpose. – Philipp Jan 6 '12 at 15:34
add comment
1 Answer
active oldest votes
These matrix equations are called Lyapunov equations and are extensively studied in control theory.
up vote 10 down vote For instance, if $A$ is Hurwitz (all eigenvalues in the left half-plane), then the unique symmetric solution of $A^TX+XA+Q$ is $$ X=\int_0^\infty e^{A^T t } Q e^{At} dt. $$
thank you!! this is exactly what I was looking for! – Philipp Jan 6 '12 at 15:19
could you please point me to a reference for the above result? thanks! – Philipp Jan 6 '12 at 15:37
Check page 230 of math.rutgers.edu/~sontag/mct.html (there is a full version of the book available online on the page). The proof is not complicated. – Federico Poloni
Jan 6 '12 at 17:32
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Not the answer you're looking for? Browse other questions tagged matrices linear-algebra or ask your own question.
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Math Forum Discussions - Visualiing Derivatives with Cubes
Date: Sep 1, 2013 12:37 PM
Author: Joe Niederberger
Subject: Visualiing Derivatives with Cubes
Its easy to visualize what's going on with the derivatives of x**2 and x**3 with the usual square and cube representations of those functions: a square can be enlarged by "building out" along two edges, a cube can be likewise by "building out" on three faces -- the "error" artifacts are the little dx corner square in the 2D case, the corner cube as well as the three edge "lines" in the 3D case. Its just a cute way of seeing where the derivatives d/dx(x**2) = 2x, and d/dx(x**3) = 3(x**2) come from.
But I don't see how to do anything similar with triangles or tetrahedrons. Perhaps Kirby will show us. Or does this simple exercise point to something a bit more fundamental than simple "cultural choice"?
Joe N
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IntroductionThe Basic Displacement Sensor, and the Design and Analysis of the Primary Conditioning CircuitPrimary Conditioning of Sensor OutputQuality Analysis of the Primary Conditioning CircuitFidelity of Primary AmplifierSensitivity AnalysisNonlinearity EffectsPractical Realisation of the Signal Conditioning AmplifierSensitivity EnhancementVoltage-to-Frequency (VFC) ConversionComponent's Selection for the AD 650 VFCSimulations, Experimental Validation, Results Presentation and DiscussionsSimulationsExperimental ValidationResult DiscussionsCost-AnalysisConclusionsLimitationsReferencesFigures and Tables
Sensors Sensors 1424-8220 Molecular Diversity Preservation International (MDPI) 10.3390/s120810820 sensors-12-10820 Article A Linear, Millimetre Displacement-to-Frequency Transducer AgeeJohn T.^1^*
PettoFour K.^2 Department of Electrical Engineering, Tshwane University of Technology, Private Bag X680, Pretoria 0001, South Africa P.O. Box 401783, Broadhurst, Gaborone, Botswana; E-Mail:
pettok@bpc.bw Author to whom correspondence should be addressed; E-Mail: ageejt@tut.ac.za; Tel.: +27-71-500-5901. 2012 06 08 2012 12 8 10820 10833 01 07 2012 21 07 2012 26 07 2012 © 2012 by the
authors; licensee MDPI, Basel, Switzerland. 2012
This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution license (http://creativecommons.org/licenses/by/3.0/).
The paper presents a novel linear, high-fidelity millimetre displacement-to-frequency transducer, based on the resistive conversion of displacement into a proportional voltage, and then frequency.
The derivation of the nonlinearity, fidelity and sensitivity of the transducer is presented. Experimental results confirm that a displacement of 0–100 mm is converted into a frequency range of 0–100
kHz, with a normalised fidelity factor of 99.91%, and a worst-case nonlinearity of less than 0.08%. Tests using laboratory standards show that a displacement of 10 mm is transduced with an accuracy
of ±0.6%, and a standard deviation of 530 Hz. Estimates included in the paper show that the transducer could cost less than 1% of existing systems for millimeter displacement measurement.
millimetre displacement displacement-to-frequency high-fidelity
Several ultrasound, optical or laser-based devices exist for the measurement of displacements larger than one metre [1–4]. The cost of the modifications required for the use of these systems for
measuring displacements in the range of a few micrometres to millimetres (submetre) is only justifiable in a few circumstances. For affordable submetre displacement measurements, capacitive and
inductive position sensors are often used. However, the frequency dependence of capacitive and inductive sensors limits their domains of application [5–7]. In fact, a comparative discourse relating
the range of displacement measurable versus the sensor recommended, could be found in [8].
In process and industrial instrumentation systems, several variables are detected using elastic sensors as primary sensing elements. Elastic sensors often generate displacements in the range of
several micrometres to millimeters, which have to be conditioned further. Table 1, derived from information available in Chapter 8 of [9], shows example applications of elastic sensors resulting in
an intermediate displacement variable. Moreover, physiological changes in biological tissues resulting from dehydration, accumulation of fluid due to disease, etc., can be studied using submetre
displacement measurements [10]. Millimeter displacement is also encountered in the analysis of the integrity of civil structures [11,12], where such measurement systems as the GPS-RTS are currently
used. A key challenge in the current systems for millimeter-displacement measurement is the high cost of acquisition of such measurement systems. Hence, there is significant motivation for the
exploration of cheaper systems for use in small displacement measurement. Moreover, the transducers most suitable for the conditioning of such small displacement signals must have high sensitivity,
high fidelity and minimum nonlinearity for acceptable accuracy of transduction. This paper presents the design, analysis and experimental validation of a submetre displacement-to-frequency
transducer. The system is based on the sensitivity of some resistive elements to displacement. Resistive sensors are relatively cheap; and their zero-order dynamics make them suitable for both static
and dynamic measurements. Unlike time-of-flight devices or phase-based measurement systems, resistive millimeter displacement transducers need be coupled physically to the displacement being
In the rest of the paper, the circuit design, analysis of the basic displacement-to-voltage converter, and the implementation of the primary conditioning amplifier circuit are presented in Section 2.
The voltage-to-frequency conversion design is presented in Section 3 of the paper. Circuit realisation, experimental results, and discussions of these results form Section 4 of the paper. Section 5
presents conclusions and the limitation of the transducer circuit. A list of references concludes the paper.
The basic displacement-to-voltage sensor is shown in Figure 1. The sensor consists of a three-terminal potentiometer of total resistance R[P], supplied by a DC voltage V[s]. The resistance between
terminals A and B of the potentiometer is directly related to the displacement d(t) (alternatively, the normalised displacement x), where 0 ≤ x = d d T ≤ 1 ; and d(t) = xd[T]. Then: E T H V s = R p x
R p o r E T H = V s x
Note that, the maximum value of E[TH] is V[s], when x = 1. The Thevenin's resistance of the equivalent sensor circuit, R[TH], is evaluated to yield [9]: R T H = R p ( 1 − x ) R p x R p ( 1 − x ) + R
p x = R P x ( 1 − x )
The sensitivity of E[TH] to the normalised displacement x is given by: S E T H x = ∂ E T H ∂ x = V s
To avoid excessive power dissipation in the resistance of the potentiometer, V[s] is usually kept small.
Consequently, the sensitivity S E T H x of this basic sensor is small. Additional conditioning is required to improve the sensitivity of the sensor. Now, the normalised sensitivity is given by: S μ E
T H x = 1 V s ∂ E T H ∂ x = 1
The equivalent circuit resulting from the connection of a primary amplifier of input resistance R[L], across terminals AB of the sensor circuit is shown in Figure 2. Loading effects tend to degrade
the performance of amplifiers. The loading effect of the conditioning circuit modifies the Thevenin's voltage to: V L = R L R L + R T H ⋅ V s x
The normalised value of this voltage is also derived to be: V L μ = V L V s = R L x R L + R P x ( 1 − x )
In the next sub-section of the paper, Equations (5) and (6) are used to analyse the quality of the displacement-to-voltage conversion amplifier, and to show any additional condition(s) that could be
imposed on the conditioning circuit to further improve the performance of the transducer.
In this subsection, the analysis of the quality of the primary signal amplification, based on Equation (5), is presented.
Fidelity is a measure of how faithfully a circuit has processed a given signal to minimize distortions. The concept of fidelity is usually used in the analysis of high frequency amplifiers. In the
current paper, the concept of fidelity is used to quantify the loading effect of the primary conditioning amplifier on the signal produced by the sensor. Now, the voltage drop due to the loading
effect of R[L] is obtained to be: Δ V = E T H − V L = R P ( x 2 − x 3 ) R L + R P ( x − x 2 ) V sor, in normalised form: Δ V μ = Δ V V s = R P ( x 2 − x 3 ) R L + R P ( x − x 2 )
The normalised fidelity factor is then given as: F μ = 1 − Δ V μ = 1 + K x ( 1 − x ) 2 1 + K x ( 1 − x )
For perfect fidelity, K = 0. Practically, this requires that: K = R P R L → 0 ; R L → ∞
The sensitivity of the conditioned output is derived to yield: S V L x = ∂ V L ∂ x = R L 2 + R L R P x 2 [ R L + R P x ( x − x 2 ) ] 2 V s
The normalised sensitivity is found to be: S V L μ x = 1 + K x 2 [ 1 + K ( x − x 2 ) ] 2
It is required to select the value of K in such a manner, as to minimize variations S V L μ x within the range of measurements.
From Equation (5)V[L] is nonlinear in x. The nonlinearity N(x) can be quantified by using: V L = m ∗ x + c ∗ + N ( x )where the linear part of V[L] is defined by the following parameters: m ∗ = V L (
x max ) − V L ( x min ) x max − x min V S c ∗ = V L ( x min ) − m ∗ x min ; x max = 1 , x min = 0 , c ∗ = 0and the nonlinearity N(x) is given as: ∴ N ( x ) = V s x − m ∗ x − c ∗ N ( x ) = V s { K x 2
( 1 − x ) 1 + K x ( 1 − x ) } N μ ( x ) = K x 2 ( 1 − x ) 1 + K x ( 1 − x )
Nonlinearity is not desirable, and is eliminated as in Equation (10). In fact, it is evident from Equations (9), (10) and (15) that K → 0 improves linearity and fidelity. This contradicts the
requirement for enhanced sensitivity as in Equation (12), for which K → ∞.The approach in this paper is to select K → 0 for fidelity and linearity enhancement; and to effect sensitivity improvement
using voltage to frequency conversion.
The practical implementation of the primary conditioning amplifier uses the summing amplifier shown in Figure 3 [13,14], with the amplified voltage given by: v 0 = ( 1 + R 2 / R 1 ) V L = ( 1 + R 2 /
R 1 ) V s 1 + K x ( 1 − x ) x = ( 1 + R 2 / R 1 ) V s 1 + K x ( 1 − x ) d ( t ) d T
An amplifier gain of 10 was used for the current work. This yields the normalised sensitivity parameter given in Equation (17).
S v 0 x = 10 ( 1 + K x 2 ) [ 1 + K ( x − x 2 ) ] 2
Table 2 summarises the parameters of the sensor and the amplifying circuit.
Combining the parameter values in Table 2 with Equations (5) and (10) yields: 0.998 x ≤ V L = V S 1 + 0.0056 x ( 1 − x ) x ≤ 1.006 x
Therefore, within an accuracy of ±0.6%: V L = x [ V ]Similarly: v 0 = 10 V L = 10 x = 10 d ( t ) d T [ V ] ; d ( t ) , d T [ m m ]and the sensitivity of the amplified voltage is given by: S v 0 d =
0.1 [ V / m m ] ; d T = 100 m m
In Section 3, we present a technique to further improve the sensitivity of the transducer, using voltage-to-frequency conversion.
As observed above, a small value of K (0.0056) was required to both minimize nonlinearity effects, and to enhance fidelity of the primary conditioning circuit. This value of K however, lowers the
sensitivity of the transduction process. Since submetre displacements can be very small, a very high sensitivity transducer is required (as shown in Table 2, ideal sensitivity required is ∞). In the
sequel, we present a voltage-to-frequency converter circuit that is used to further enhance the sensitivity of the developed transducer.
Apart from sensitivity enhancement, the conversion of v[0] into a frequency signal has several other advantages, including: high noise immunity, high output power, wide dynamic range, and ease of
interfacing with digital data acquisition systems. Table 3 shows key values of v[0] and their corresponding frequency representations.
The linear relationship between v[0] and frequency in Table 3 is expressed mathematically as: f 0 = 10000 v 0 [ H z ]
Applying Equation (20) in Equation (22) we obtain: f 0 = 1000 d ( t ) [ H z ] ; d ( t ) [ m m ]
It is evident from Equation (23) that: d ( t ) = f 0 1000 [ m m ] ; f 0 [ H z ]
The AD 650 voltage-to-frequency converter (VFC) was used for the implementation of the displacement-to-frequency conversion circuit satisfying Equation (23). The pin layout of the AD 650 VFC is
obtained from the manufacturer's manual for the device [15]. The selection of components for the VFC circuit is presented in the sequel.
For the AD 650, only four component values must be selected by the user [15]. Using the manufacturer's notation, these are the input resistance R[IN], the timing capacitor C[OS], the logic resistor R
[2] and the integration capacitor C[INT]. The first two are determined by the input voltage range and full-scale frequency. Additional relationship between R[IN] and C[OS] is provided through graphs
obtainable in [15]. Sample design for a maximum frequency of 100 KHZ in the data sheet of the AD 650 VFC used R[IN] = 40 kΩ and this has been adopted for the realization in this study. Table 4
summarizes the components used for the design of the VFC circuit, with C[INT] calculated using the equation: C INT = 10 − 4 F / sec f MAX ≥ 1000 p F
The frequency conversion improves the sensitivity of the transducer from the value given by Equation (17) to: S v 0 H z = 100000 − 0 ( 100 − 0 ) = 10 3 bits / m mand has also improved the resolution
to = 10^−3 mm/bit.
The indices assessing the quality of the transducer were evaluated by simulation. The rest of the results were obtained through experimental measurements.
MATLAB simulation of the transducer fidelity, sensitivity and nonlinearity, based on Equations (9), (12) and (15) is presented in Figures 4–6.
The experimental setup is shown in Figure 7. For the experiments, a slide wire potentiometer was used as the submeter displacement sensor. It had a maximum displacement d[T] = 100 mm = 10^−1 m, and a
total resistance of 11.2 kΩ (instead of the design maximum resistance of 10 kΩ). The potentiometer was supplied by a 1volt DC supply. The Thevenin voltage of the sensor, as a function of
displacement, is shown in Figure 8. A plot of the amplified sensor voltage as function of detected displacement is shown in Figure 9. The overall displacement-to-frequency transduction is shown in
Figure 10. For the analysis of the accuracy and precision of transducing displacement inputs into frequency, repeated measurements of 10 mm displacement were undertaken. The results are shown in
Figures 11 and 12.
The simulated fidelity results presented in Figure 4 shows that, the conditioning of the sensor signal by the primary amplifier was undertaken with at least 99.91% fidelity. Figure 5 shows that the
introduction of the conditioning circuit slightly reduced the normalized input sensitivity ( S μ E t h x ) to a value less than unity (0.988). As shown in Figure 6, the choice of K = 0.0056 has
reduced the worst case nonlinearity to 0.18% < 4%, as typically allowed in instrumentation [9]. The experimental measurements, confirmed the linearity between the displacement and the
displacement-dependent voltage E[TH] as in Figure 8. Figure 9 shows that, the connection of the voltage amplifier has introduced nearly 0.22% nonlinearity (obtained from the linear correlation
coefficient); also slightly reducing the normalized sensitivity from a value of unity to 0.9978. Experimental results of the overall displacement-to-frequency conversion process are shown in Figure
10. It is evident there-from, that the millimetre-to-frequency converter has an overall linearity to 99.92%, or a nonlinearity of 0.08%. Results from the precision analysis of the
displacement-to-frequency transducer are shown in Figures 11 and 12. It is evident from these, that for a 10 mm displacement, a mean measurement of 10,062 Hz (for 100,000 Hz) was obtained, giving a
transducer accuracy of ±0.62%; the standard deviation of the measurements was 530 Hz.
The reported millimeter-to-frequency transducer consists of one resistive sensor, five resistors, two capacitors, two operational amplifiers and one AD 650 voltage-to-frequency converter. Table 5
shows the cost estimation for the new transducer based on average component cost, and 30% device production fee. Calculations are shown for two sensor types: potentiometers and strain gauges.
To put the above costs in perspective, Table 6 compares the cost of the reported transducer with those of existing displacement sensors.
It is evident from Table 6, that the reported transducer has a very significant financial advantage over several existing systems for displacement measurements.
It is concluded that a cheap, linear, millimetre displacement-to-frequency transducer with both high sensitivity and high fidelity has been successfully realised.
The design sensor resistance of 10 kΩ was not available. A sensor of total resistance of 11.2 kΩ was used instead. Whereas this larger resistance value did not directly affect the accurate
performance of the transducer, it was observed that, the maximum output frequency was 120 kHz (instead of the design maximum frequency of 100 kHz). Temperature variations constitute a significant
random impact on sensor performance. Temperature effects have not yet being characterized. The effect of supply voltage variation is also still under investigation. Test measurements were undertaken
using laboratory standards. Traceability of accuracy shall be undertaken in subsequent development, using facilities at a national metrology centre.
SongH.X.WangX.D.MaL.Q.CaiM.Z.CaoT.Z.Design and performance analysis of a laser displacement sensor based on Position Sensitive Detector (PSD)200648217222 RiedijkF.R.SmithT.HuijsingJ.H.An integrated
optical position-sensitive detector with digital output and error correction19944023724210.1016/0924-4247(94)87010-1 HofstetterD.ZappeH.P.DändlikerR.Optical displacement measurement with GaAs/
AlGaAs-based monolithically integrated michelson interferometers19971566367010.1109/50.566688 McCarthyM.DuffP.MullerH.L.RandellC.Accessible ultrasonic positioning200658693 AkioK.Design and
characterization of nano-displacement sensor with high-frequency oscillators20112011360173 YangH.LiR.WeiQ.LiuJ.The study of high accuracy capacitive displacement sensor used in non-contact precision
displacement measurementProceedings of the 9th International Conference on Electronic Measurement and Instruments (ICEMI '09)Beijing, China16–19 August 2009164168 RahalM.DemosthenousA.
D.JiangD.PalD.A signal conditioner for high-frequency inductive position sensorsProceedings of the 2008 International Conference on Microelectronics (ICM 2008)Sharjah, UAE14–17 December 2008118122
ShiehJ.HuberJ.E.FleckN.A.AshbyM.F.The selection of sensors20014646150410.1016/S0079-6425(00)00011-6 BentleyJ.P.Sensing Elements3rd ed.Pearson Education LimitedEdinburgh, UK1995135177 LionK.S.Electric
displacement transducers for physiological investigations1953129830610.1093/rheumatology/1.8.298 CasciatiF.FugginiC.Engineering vibration monitoring by GPS: Long duration records2009845946710.1007/
s11803-009-9058-8 PsimoulisP.PytharouliS.KarambalisD.StirosS.Potential of global positioning system (GPS) to measure frequencies of oscillations of engineering structures200831860662310.1016/
j.jsv.2008.04.036 GeorgeB.R.1st ed.John Wiley and SonsNew York, NY, USA1993 ChirlianP.M.2nd ed.John Wiley and SonsNew York, NY, USA1987 Manufacturer's Datasheet for the AD 650 Voltage-to-Frequency
ConverterAvailable online: http://www.datasheetcatalog.com (accessed on 25 December 2011)
Arrangement of potentiometer-type displacement sensor.
Equivalent circuit of sensor with primary conditioning amplifier.
Summing amplifier used as the primary conditioning circuit.
Fidelity of primary voltage conditioning amplifier.
Sensitivity analysis in displacement-to-voltage conversion.
Analysis of nonlinearity in displacement to voltage conversion.
Experimental setup of transducer.
Basic sub-meter-to-voltage conversion.
Performance of voltage amplifier.
Displacement to frequency transduction.
Sample measurements from transducer for precision analysis.
Equivalent millimeter output from frequency measurements.
Example of elastic sensors producing submtre displacement as output [9].
Elastic Sensor Primary Variable
Strain gauge Force
Parallel plate capacitor Pressure
Diaphragms Pressure
Bellows Pressure
Bourdon tubes Pressure
Load cells Pressure
Cantilever sensors Force
Cylindrical shafts Torque
Proving ring Force
System parameters.
Parameter Design Value Ideal/Used Value
d[T] 100 mm
R[P] 10 kΩ 11.2 kΩ
R[L] 2 MΩ ∞
R[1] 20 kΩ
R[2] 180 kΩ
V[s] 1.0 V
K = R[P]/R[L] 0.0056 0
N[μ](x) 0≤ N[μ](x) ≤ 0.00083 0
F[μ] 0.9992 ≤ F[μ] ≤ 1.000 1
S V L μ x 0.9981 ≤ S V L μ x ≤ 1.0056 ∞
Voltage-to-frequency table.
d (mm) x v[0] (V) Frequency (kHz)
Parameters of the voltage-to-frequency converter.
Parameter Value
f[max] 100 kHz
R[in] 40 kΩ
C[INT] 1,000 pF
C[OS] 330 pF
R[2] 1.75 kΩ
Cost analysis per unit of the millimeter-to-frequency transducer.
Potentiometer sensor: $11.0/unit $48.70
Components 5× Resistors: $7.50 2× Capacitors: $6.0 2× Op Amp: $2.0 1× AD 650: $11.00 Labour costs at 30% Total transducer cost
Strain gauge: $3.50/unit $39.00
Cost comparison with some existing position sensors
Sensor Cost Source of Information
Potentiometer millimeter-to-frequency converter $39,00 Table 5 in the paper
Strain gauge type Millimeter-to-frequency transducer $48.7.00 Table 5 in the paper
SwissRanger 4000 (SR4000) $9,000 http://www.hizook.com/blog/2010/03/28/low-cost-depth-cameras-aka-ranging-cameras-or-rgb-d-cameras-emerge-2010
PMD Technologies CamCube 2.0 $12,000 http://www.hizook.com/blog/2010/03/28/low-cost-depth-cameras-aka-ranging-cameras-or-rgb-d-cameras-emerge-2010
Hewlett Packard model 5525A Laser system $11,500 http://www.n4mw.com/hp5526/hple.htm
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4 Oil And Gas Stocks Undervalued To The Graham Number
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Graham Number = SQRT(22.5 x TTM EPS x MRQ BVPS)
The equation assumes that P/E should not be higher than 15, and P/BV should not be higher than 1.5. Stocks trading well below their Graham Number may be undervalued.
For an interactive version of this chart, click on the image below. Tool provided by Kapitall. Login to access free research tools, share practice portfolios, and more.
Do you think these stocks should be trading higher? Use this list as a starting point for your own analysis.
1. C&J Energy Services, Inc. (CJES): Provides specialty equipment services for oil and natural gas exploration and production companies in the Texas, Louisiana, and Oklahoma regions of the United
States. Market cap at $1.22B, most recent closing price at $23.03. Diluted TTM earnings per share at 3.82, and a MRQ book value per share value at 10.61, implies a Graham Number fair value = sqrt
(22.5*3.82*10.61) = $30.20. Based on the stock's price at $23.02, this implies a potential upside of 31.18% from current levels.
2. Gulfmark Offshore, Inc. (GLF): Offers marine specialty services to offshore oil and natural gas drilling rigs and platforms. Market cap at $924.58M, most recent closing price at $34.32. Diluted
TTM earnings per share at 1.82, and a MRQ book value per share value at 39.4, implies a Graham Number fair value = sqrt(22.5*1.82*39.4) = $40.17. Based on the stock's price at $34.65, this implies a
potential upside of 15.92% from current levels.
3. Hess Corporation (HES): Operates as an integrated energy company. Market cap at $23.18B, most recent closing price at $67.88. Diluted TTM earnings per share at 4.46, and a MRQ book value per
share value at 60.67, implies a Graham Number fair value = sqrt(22.5*4.46*60.67) = $78.03. Based on the stock's price at $58.9, this implies a potential upside of 32.47% from current levels.
4. Tesoro Corporation (TSO): Engages in refining and marketing petroleum products in the United States. Market cap at $6.75B, most recent closing price at $48.09. Diluted TTM earnings per share at
4.19, and a MRQ book value per share value at 31.45, implies a Graham Number fair value = sqrt(22.5*4.19*31.45) = $54.45. Based on the stock's price at $44.58, this implies a potential upside of
22.14% from current levels.
|
{"url":"http://business-news.thestreet.com/deming-headlight/story/4-oil-and-gas-stocks-undervalued-to-the-graham-number/11829283","timestamp":"2014-04-17T04:21:08Z","content_type":null,"content_length":"20450","record_id":"<urn:uuid:cc2c4d36-4ba0-45bd-96ad-911b008b5302>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00514-ip-10-147-4-33.ec2.internal.warc.gz"}
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14 Hours ago
cleared ibm 2 rounds
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18 Hours ago
Thanks to m4maths.I got placed in IBM.Awsome work.Best of luck.
Lekshmi Narasimman MN
6 Days ago
Thanks ton for this site . This site is my main reason for clearing cts written which happend on 5/4/2014 in chennai . Tommorrw i have my interview. Hope i will tel u all a good news :)
Thanks to almighty too :) !!
abhinay yadav
11 Days ago
thank you M4maths for such awesome collection of questions. last month i got placed in techMahindra. i prepared for written from this site, many question were exactly same as given here.
bcz of practice i finished my written test 15 minutes before and got it.
thanx allot for such noble work...
16 Days ago
coz of this site i cud clear IBM's apti nd finally got placed in tcs
thanx m4maths...u r a wonderful site :)
18 Days ago
thank u m4maths and all its user for posting gud and sensible answers.
Nilesh singh
21 Days ago
finally selected in TCS. thanks m4maths
23 Days ago
Thank you team m4maths.Successfully placed in TCS.
Deepika Maurya
23 Days ago
Thank you so much m4maths.. I cleared the written of IBM.. :) very good site.. thumps up !!
Rimi Das
1 Month ago
Thanks to m4maths I got selected in Tech Mahindra.I was preparing for TCS 1st round since last month.Got interview call letter from there also...Really m4maths is the best site for placement
Stephen raj
1 Month ago
prepare from r.s.aggarwal verbal and non verbal reasoning and previous year questions from m4maths,indiabix and chetanas forum.u can crack it.
Stephen raj
1 Month ago
Thanks to m4maths:)
cracked infosys:)
1 Month ago
i have been Selected in Tech Mahindra.
All the quanti & reasoning questions are common from the placement papers of m4maths.
So a big thanks to m4maths team & the people who shares the placement papers.
Amit Das
1 Month ago
I got selected for interview in TCS.Thank you very much m4maths.com.
1 Month ago
I got placed in TCS :)
Thanks a lot m4maths :)
Syed Ishtiaq
1 Month ago
An Awesome site for TCS.
Cleared the aptitude.
1 Month ago
I successfully cleared TCS aptitude test held on 8th march 2014.Thanks a lot m4maths.com
plz guide for the technical round.
mounika devi mamidibathula
1 Month ago
got placed in IBM..
this site is very useful, many questions repeated..
thanks alot to m4maths.com
Anisha Lakhmani
1 Month ago
I got placed at infosys.......thanx to m4maths.com.......a awesum site......
Anisha Lakhmani
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2 Months ago
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Surya Narayana K
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I successfully cleared TCS aptitude test.Thanks a lot m4maths.com.
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Got Placed In Infosys...
Thanks of m4maths....
3 Months ago
iam not placed in TCS...........bt still m4maths is a good site.
4 Months ago
Thanx to m4 maths, because of that i able to crack aptitude test and now i am a part of TCS. This site is best for the preparation of placement papers.Thanks a lotttttt............
4 Months ago
THANKS a lot m4maths. Me and my 2 other roomies cleared the tcs aptitude with the help of this site.Some of the questions in apti are exactly same which i answered without even reading the whole
question completely.. gr8 work m4maths.. keep it up.
5 Months ago
m4maths is one of the main reason I cleared TCS aptitude. In TCS few questions will be repeated from previous year aptis and few questions will be repeated from the latest campus drives that happened
in various other colleges. So to crack TCS apti its enough to learn some basic concepts from famous apti books and follow all the TCS questions posted in m4maths. This is not only for TCS but for all
other companies too. According to me m4maths is best site for clearing apti. Kuddos to the creator of m4maths :)
5 Months ago
THANKS A LOT TO M4MATHS.due to m4maths today i am the part of TCS now.got offer letter now.
5 Months ago
Hai friends, I got placed in L&T INFOTECH and i m visiting this website for the past 4 months.Solving placemetn puzzles from this website helped me a lot and 1000000000000s of thanks to this
website.this website also encouraged me to solve puzzles.follw the updates to clear maths aps ,its very easy yar, surely v can crack it if v follow this website.
5 Months ago
2 days before i cleared written test just because of m4maths.com.thanks a lot for this community.
6 Months ago
thanks for m4maths!!! bcz of which i cleared apti of infosys today.
6 Months ago
Today my written test of TCS was completed.I answered many of the questions without reading entire question.Because i am one of the member in the m4maths.
No words to praise m4maths.so i simply said thanks a lot.
7 Months ago
I am very grateful to m4maths. It is a great site i have accidentally logged on when i was searching for an answer for a tricky maths puzzle. It heped me greatly and i am very proud to say that I
have cracked the written test of tech-mahindra with the help of this site. Thankyou sooo much to the admins of this site and also to all members who solve any tricky puzzle very easily making people
like us to be successful. Thanks a lotttt
Abhishek Ranjan
7 Months ago
me & my rooom-mate have practiced alot frm dis site TO QUALIFY TCS written test.both of us got placed in TCS :)
do practice n u'll surely succeed :)
Sandhya Pallapu
1 year ago
Hai friends! this site is very helpful....i prepared for TCS campus placements from this site...and today I m proud to say that I m part of TCS family now.....dis site helped me a lot in achieving
this...thanks to M4MATHS!
vivek singh
2 years ago
I cracked my first campus TCS in November 2011...i convey my heartly thanks to all the members of m4maths community who directly or indirectly helped me to get through TCS......special thanks to
admin for creating such a superb community
Manish Raj
2 years ago
this is important site for any one ,it changes my life...today i am part of tcs only because of M4ATHS.PUZZLE
Asif Neyaz
2 years ago
Thanku M4maths..due to u only, imade to TCS :D test on sep 15.
Harini Reddy
2 years ago
Big thanks to m4maths.com.
I cracked TCS..The solutions given were very helpful!!!
2 years ago
HI everyone ,
me and my friends vish,sube,shaf placed in TCS... its becoz of m4maths only .. thanks a lot..this is the wonderful website.. unless your help we might not have been able to place in TCS... and thanks
to all the users who clearly solved the problems.. im very greatful to you :)
2 years ago
Really thanks to m4maths I learned a lot... If you were not there I might not have been able to crack TCS.. love this site hope it's reputation grows exponentially...
2 years ago
Hello friends .I was selected in TCS. Thanx to M4Maths to crack apti. and my hearthly wishes that
the success rate of M4Math grow exponentially.
Again Thanx for all support given by M4Math during
my preparation for TCS.
and Best of LUCK for all students for their preparation.
2 years ago
thanks to M4MATHS..got selected in TCS..thanks for providing solutions to TCS puzzles :)
2 years ago
thousands of thnx to m4maths...
got selected in tcs for u only...
u were the only guide n i hv nvr done group study for TCS really feeling great...
thnx to all the users n team of m4maths...
3 cheers for m4maths
2 years ago
thousands of thnx to m4maths...
got selected in tcs for u only...
u were the only guide n i hv nvr done group study for TCS really feeling great...
thnx to all the users n team of m4maths...
3 cheers for m4maths
2 years ago
Thank U ...I'm placed in TCS.....
Continue this g8 work
2 years ago
thank you m4maths.com for providing a web portal like this.Because of you only i got placed in TCS,driven on 26/8/2011 in oncampus
raghu nandan
2 years ago
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V.V.Ravi Teja
3 years ago
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3 years ago
got placed in TCS ...
thanku m4maths...
Amulya Punjabi
3 years ago
Hi All,
Today my result for TCS apti was declared nd i cleared it successfully...It was only due to m4maths...not only me my all frnds are able to crack it only wid the help of m4maths.......it's just an
osum site as well as a sure shot guide to TCS apti......Pls let me know wt can be asked in the interview by MBA students.
Anusha Alva
3 years ago
a big thnks to this site...got placed in TCS!!!!!!
Oindrila Majumder
3 years ago
thanks a lot m4math.. placed in TCS
Pushpesh Kashyap
3 years ago
superb site, i cracked tcs
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3 years ago
Great site..........got Placed in TCS...........thanx a lot............do not mug up the sol'n try to understand.....its AWESOME.........
Gautam Kumar
3 years ago
it was really useful 4 me.................n finally i managed to get through TCS
Karthik Sr Sr
3 years ago
i like to thank m4maths, it was very useful and i got placed in tcs
venkaiaha 14 Hours ago cleared ibm 2 rounds thank you m4maths.com
Triveni 18 Hours ago Thanks to m4maths.I got placed in IBM.Awsome work.Best of luck.
Lekshmi Narasimman MN 6 Days ago Thanks ton for this site . This site is my main reason for clearing cts written which happend on 5/4/2014 in chennai . Tommorrw i have my interview. Hope i will tel u
all a good news :) Thanks to almighty too :) !!
abhinay yadav 11 Days ago thank you M4maths for such awesome collection of questions. last month i got placed in techMahindra. i prepared for written from this site, many question were exactly same
as given here. bcz of practice i finished my written test 15 minutes before and got it. thanx allot for such noble work...
manasi 16 Days ago coz of this site i cud clear IBM's apti nd finally got placed in tcs thanx m4maths...u r a wonderful site :)
arnold 18 Days ago thank u m4maths and all its user for posting gud and sensible answers.
Nilesh singh 21 Days ago finally selected in TCS. thanks m4maths
MUDIT 23 Days ago Thank you team m4maths.Successfully placed in TCS.
Deepika Maurya 23 Days ago Thank you so much m4maths.. I cleared the written of IBM.. :) very good site.. thumps up !!
Rimi Das 1 Month ago Thanks to m4maths I got selected in Tech Mahindra.I was preparing for TCS 1st round since last month.Got interview call letter from there also...Really m4maths is the best site
for placement preparation...
Stephen raj 1 Month ago prepare from r.s.aggarwal verbal and non verbal reasoning and previous year questions from m4maths,indiabix and chetanas forum.u can crack it.
Stephen raj 1 Month ago Thanks to m4maths:) cracked infosys:)
Ranadip 1 Month ago i have been Selected in Tech Mahindra. All the quanti & reasoning questions are common from the placement papers of m4maths. So a big thanks to m4maths team & the people who
shares the placement papers.
Amit Das 1 Month ago I got selected for interview in TCS.Thank you very much m4maths.com.
PRAVEEN K H 1 Month ago I got placed in TCS :) Thanks a lot m4maths :)
Syed Ishtiaq 1 Month ago An Awesome site for TCS. Cleared the aptitude.
sara 1 Month ago I successfully cleared TCS aptitude test held on 8th march 2014.Thanks a lot m4maths.com plz guide for the technical round.
mounika devi mamidibathula 1 Month ago got placed in IBM.. this site is very useful, many questions repeated.. thanks alot to m4maths.com
Anisha Lakhmani 1 Month ago I got placed at infosys.......thanx to m4maths.com.......a awesum site......
Kusuma Saddala 1 Month ago Thanks to m4maths, i have place at IBM on feb 8th of this month
sangeetha 2 Months ago thanks to m4 maths because of this i clear csc written test
mahima srivastava 2 Months ago Placed at IBM. Thanks to m4maths. This site is really very helpful. 95% questions were from this site.
Surya Narayana K 2 Months ago I successfully cleared TCS aptitude test.Thanks a lot m4maths.com.
prashant gaurav 2 Months ago Got Placed In Infosys... Thanks of m4maths....
vishal 3 Months ago iam not placed in TCS...........bt still m4maths is a good site.
sameer 4 Months ago Thanx to m4 maths, because of that i able to crack aptitude test and now i am a part of TCS. This site is best for the preparation of placement papers.Thanks a
Sonali 4 Months ago THANKS a lot m4maths. Me and my 2 other roomies cleared the tcs aptitude with the help of this site.Some of the questions in apti are exactly same which i answered without even
reading the whole question completely.. gr8 work m4maths.. keep it up.
Kumar 5 Months ago m4maths is one of the main reason I cleared TCS aptitude. In TCS few questions will be repeated from previous year aptis and few questions will be repeated from the latest campus
drives that happened in various other colleges. So to crack TCS apti its enough to learn some basic concepts from famous apti books and follow all the TCS questions posted in m4maths. This is not
only for TCS but for all other companies too. According to me m4maths is best site for clearing apti. Kuddos to the creator of m4maths :)
YASWANT KUMAR CHAUDHARY 5 Months ago THANKS A LOT TO M4MATHS.due to m4maths today i am the part of TCS now.got offer letter now.
ANGELIN ALFRED 5 Months ago Hai friends, I got placed in L&T INFOTECH and i m visiting this website for the past 4 months.Solving placemetn puzzles from this website helped me a lot and
1000000000000s of thanks to this website.this website also encouraged me to solve puzzles.follw the updates to clear maths aps ,its very easy yar, surely v can crack it if v follow this website.
MALLIKARJUN ULCHALA 5 Months ago 2 days before i cleared written test just because of m4maths.com.thanks a lot for this community.
Madhuri 6 Months ago thanks for m4maths!!! bcz of which i cleared apti of infosys today.
DEVARAJU 6 Months ago Today my written test of TCS was completed.I answered many of the questions without reading entire question.Because i am one of the member in the m4maths. No words to praise
m4maths.so i simply said thanks a lot.
PRATHYUSHA BSN 7 Months ago I am very grateful to m4maths. It is a great site i have accidentally logged on when i was searching for an answer for a tricky maths puzzle. It heped me greatly and i am
very proud to say that I have cracked the written test of tech-mahindra with the help of this site. Thankyou sooo much to the admins of this site and also to all members who solve any tricky puzzle
very easily making people like us to be successful. Thanks a lotttt
Abhishek Ranjan 7 Months ago me & my rooom-mate have practiced alot frm dis site TO QUALIFY TCS written test.both of us got placed in TCS :) IT'S VERY VERY VERY HELPFUL N IMPORTANT SITE. do practice
n u'll surely succeed :)
Sandhya Pallapu 1 year ago Hai friends! this site is very helpful....i prepared for TCS campus placements from this site...and today I m proud to say that I m part of TCS family now.....dis site
helped me a lot in achieving this...thanks to M4MATHS!
vivek singh 2 years ago I cracked my first campus TCS in November 2011...i convey my heartly thanks to all the members of m4maths community who directly or indirectly helped me to get through
TCS......special thanks to admin for creating such a superb community
Manish Raj 2 years ago this is important site for any one ,it changes my life...today i am part of tcs only because of M4ATHS.PUZZLE
Asif Neyaz 2 years ago Thanku M4maths..due to u only, imade to TCS :D test on sep 15.
Harini Reddy 2 years ago Big thanks to m4maths.com. I cracked TCS..The solutions given were very helpful!!!
portia 2 years ago HI everyone , me and my friends vish,sube,shaf placed in TCS... its becoz of m4maths only .. thanks a lot..this is the wonderful website.. unless your help we might not have been
able to place in TCS... and thanks to all the users who clearly solved the problems.. im very greatful to you :)
vasanthi 2 years ago Really thanks to m4maths I learned a lot... If you were not there I might not have been able to crack TCS.. love this site hope it's reputation grows exponentially...
vijay 2 years ago Hello friends .I was selected in TCS. Thanx to M4Maths to crack apti. and my hearthly wishes that the success rate of M4Math grow exponentially. Again Thanx for all support given by
M4Math during my preparation for TCS. and Best of LUCK for all students for their preparation.
maheswari 2 years ago thanks to M4MATHS..got selected in TCS..thanks for providing solutions to TCS puzzles :)
GIRISH 2 years ago thousands of thnx to m4maths... got selected in tcs for u only... u were the only guide n i hv nvr done group study for TCS really feeling great... thnx to all the users n team of
m4maths... 3 cheers for m4maths
girish 2 years ago thousands of thnx to m4maths... got selected in tcs for u only... u were the only guide n i hv nvr done group study for TCS really feeling great... thnx to all the users n team of
m4maths... 3 cheers for m4maths
Aswath 2 years ago Thank U ...I'm placed in TCS..... Continue this g8 work
JYOTHI 2 years ago thank you m4maths.com for providing a web portal like this.Because of you only i got placed in TCS,driven on 26/8/2011 in oncampus
raghu nandan 2 years ago thanks a lot m4maths cracked TCS written n results are to be announced...is only coz of u... :)
V.V.Ravi Teja 3 years ago thank u m4maths because of you and my co people who solved some complex problems for me...why because due to this only i got placed in tcs and hcl also........
Veer Bahadur Gupta 3 years ago got placed in TCS ... thanku m4maths...
Amulya Punjabi 3 years ago Hi All, Today my result for TCS apti was declared nd i cleared it successfully...It was only due to m4maths...not only me my all frnds are able to crack it only wid the
help of m4maths.......it's just an osum site as well as a sure shot guide to TCS apti......Pls let me know wt can be asked in the interview by MBA students.
Anusha Alva 3 years ago a big thnks to this site...got placed in TCS!!!!!!
Oindrila Majumder 3 years ago thanks a lot m4math.. placed in TCS
Pushpesh Kashyap 3 years ago superb site, i cracked tcs
Saurabh Bamnia 3 years ago Great site..........got Placed in TCS...........thanx a lot............do not mug up the sol'n try to understand.....its AWESOME.........
Gautam Kumar 3 years ago it was really useful 4 me.................n finally i managed to get through TCS
Karthik Sr Sr 3 years ago i like to thank m4maths, it was very useful and i got placed in tcs
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"Pure mathematics is the world's best game. It is more absorbing than chess, more of a gamble than poker, and lasts longer than Monopoly. It's free. It can be played anywhere - Archimedes did it in a
bathtub." Richard J. Trudeau, Dots and Lines
"Do not worry about your difficulties in mathematics, I assure you that mine are greater." Albert Einstein
"In mathematics the art of proposing a question must be held of higher value than solving it." Georg Cantor
""Play the game with attitude, incase if you loose you will win the heart of the spectators and opponents"" Sandeep Upadhyay
"You may be an engineer if your idea of good interpersonal communication means getting the decimal point in the right place." Unknown
"It is easier to square the circle than to get round a mathematician." Augustus de Morgan
"my math textbook is my role model. no matter how many problems it has it never sucide's." prabhpreet singh
Latest Placement Puzzle (More)
"A man walked a certain distance south and then the same distance plus 7km due west.He is now 13km from his starting point.What are the distances south and west that he walked?"."
UnsolvedAsked In: TCS
"appitude questiosn
UnsolvedAsked In: Tech Mahindra
"how+much=power then p+o+w+e+e=?"
UnsolvedAsked In: infosys
"Pure mathematics is the world's best game. It is more absorbing than chess, more of a gamble than poker, and lasts longer than Monopoly. It's free. It can be played anywhere - Archimedes did it in a
bathtub." Richard J. Trudeau, Dots and Lines
"Do not worry about your difficulties in mathematics, I assure you that mine are greater." Albert Einstein
"In mathematics the art of proposing a question must be held of higher value than solving it." Georg Cantor
""Play the game with attitude, incase if you loose you will win the heart of the spectators and opponents"" Sandeep Upadhyay
"You may be an engineer if your idea of good interpersonal communication means getting the decimal point in the right place." Unknown
"It is easier to square the circle than to get round a mathematician." Augustus de Morgan
"my math textbook is my role model. no matter how many problems it has it never sucide's." prabhpreet singh
"A man walked a certain distance south and then the same distance plus 7km due west.He is now 13km from his starting point.What are the distances south and west that he walked?"." UnsolvedAsked In:
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Order Theory Category Theory
A selection of articles related to order theory category theory.
Original articles from our library related to the Order Theory Category Theory. See Table of Contents for further available material (downloadable resources) on Order Theory Category Theory.
Order Theory Category Theory is described in multiple online sources, as addition to our editors' articles, see section below for printable documents, Order Theory Category Theory books and related
Suggested Pdf Resources
Lawvere went on in [13J to formulate a (first-order) theory whose objects are conceived to be arbitrary categories and functors between them.
In these terms, the intended question really is “Does category theory pro- ..
Suggested Web Resources
Pages in category "Order theory". The following 141 pages are in this category, out of 141 total. This list may not reflect recent changes (learn more).
Functions between orders become functors between categories.
Feb 25, 2010 Eilenberg & Mac Lane invented category theory precisely in order to clarify and compare these connections.
Category: Group Theory. This post started out as a response to a question in the comments of my last post on groupoids.
Great care has been taken to prepare the information on this page. Elements of the content come from factual and lexical knowledge databases, realmagick.com library and third-party sources. We
appreciate your suggestions and comments on further improvements of the site.
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Optimization and integral help?
April 17th 2012, 11:46 AM
Optimization and integral help?
Hi guys, I'm trying to study for my exam from these practice problems and I don't know how to start. I was wondering if I can get some help. Thank you so much!
1. Three friends want to share a 12-inch diameter pizza equally by exactly only two parallel cuts. How far from the center must the cut be?
2. You are at the outermost point of a circular lake of radius 1mile. Your plan is to swim a straight course to another point on the shore of the lake, then jog to the northermost point. You can
swim 75% as fast as you can jog. Under these conditions, (a) find the route you must take to maximize the time taken for this trip and find the time in minutes. (b) find the route you must take
to minimize the time this trip will take, and find the time in minutes.
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Fourier series examples
Consider the waveforms that you can generate with the following applet:
All of these are periodic. The sinusoid has the least interesting timbre, suggesting that it will have the simplest harmonic structure. The following applet shows the Fourier series coefficients for
each waveform.
The exact waveform is shown in red, and an approximation in blue. The right edge of each red bar aligns with the scale at the bottom, so the fundamental, corresponding to the first red bar, is at
125Hz, with is one over the period of 8 milliseconds.
The approximation obtained by summing selected terms from the Fourier series. You can control which terms are used through the checkboxes on the right, but only up to 16 terms can be included in the
The Fourier series coefficients are shown on the plot labeled "Frequency domain". The phase is not shown, but rather only A[k]. The components you select are shown in blue. This lower plot is called
a frequency domain representation of the waveform because it describes the waveform in terms of its frequency components.
Notice the following about these plots:
• There is no constant term. The average value in the time domain is zero.
• The fundamental is 125 Hz.
• The first harmonic, which would be at 250 Hz, is absent for all but the sawtooth.
• The even Fourier series coefficients (k = 0, 2, 4, ...) are all absent for all but the sawtooth. This subtle property is due to the symmetry of waveforms (except for the sawtooth, which is not
• The coefficients become small quickly for the triangle wave, but not for the square wave or the sawtooth. Intuitively, the sharp discontinuities in the square wave and the sawtooth imply
relatively large high frequency components.
• For the square wave, the peak error of the approximation (the maximum difference between the red and blue curves) does not appear to decrease as the number of terms in the approximation is
increased. This is known as Gibb's phenomenon, and indeed, the square wave is one that cannot be exactly described with a Fourier series. The peak error does not decrease to zero as the number of
terms in the Fourier series is increased to infinity.
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[FOM] On Foundational Thinking 1
Harvey Friedman friedman at math.ohio-state.edu
Tue Jan 20 01:17:00 EST 2004
Foundational thinking has been an essential component in many if not most of
the greatest intellectual advances throughout history. Certainly in
Aristotle, Plato, Leibniz, Frege, Godel, Turing, Newton, Einstein, Darwin,
However, such greatest intellectual advances occur with sufficient
infrequency, that despite the essential role of foundational thinking,
foundational thinking as such has no clearly defined place in contemporary
academic life.
In particular, researchers generally do not deliberately apply foundational
thinking on a systematic basis in their research, nor do they generally
highlight the use of foundational thinking, nor do they teach foundational
thinking to students, nor do they explicitly discuss the role of
foundational thinking in the evolution of their subjects.
Foundational thinking is most explicitly well developed and highlighted in
connection with research in the foundations of mathematics. Here the various
deep and surprising results, going back over a Century, with roots in
antiquity, showcase just how powerful and effective and clarifying
foundational thinking can be.
(To avoid confusion, I draw a distinction between foundations of mathematics
and mathematical logic. The latter consists of various mathematical spinoffs
from foundations of mathematics, where one deemphasizes foundational
thinking, and emphasizes mathematical adventures, including connections with
various branches of mathematics.)
I believe that the deliberate, explicit, systematic, and imaginative use of
foundational thinking
**in ANY context of depth where there is a search for systematic knowledge
of high quality, or creation of artificial systems of great utility**
will have revolutionary effect, not only in research, but in education.
Note that I have stated ** broadly enough to go far beyond science, into
engineering, and into, e.g., economics and law (economic systems and legal
systems), as well as technology.
So what is new about this kind of wild enthusiasm for foundational thinking?
After all, great philosophers like Leibniz made similar or related
predictions, and we still find ourselves in 2004 in a deeply flawed academic
culture in which foundational thinking as such is virtually invisible; and
with unnatural, counterproductive, and wholly inappropriate sharp divisions
between "disciplines", making it all the harder for change to come about. In
fact, perhaps making it all the harder for great intellectual advances to
come about.
One thing that is new about this wild enthusiasm for foundational thinking,
over what some of the great philosophers said in the past, is that
***we now have that singularly great example of explicit systematic
foundational thinking TO LEARN FROM. That is, f.o.m.***
Great philosophers of the past such as Leibniz obviously did not have access
to this.
I think it is useful to quote Godel on Leibniz, from
K. Godel, Russell's mathematical logic, 1944, in: Collected Works, vol. II.
"[Mathematical logic] On the one hand, it is a section of mathematics ... On
the other hand, it is a science prior to all others, which contains the
ideas and principles underlying all sciences. It was in this second sense
that mathematical logic was first conceived by Leibniz in his
Characteristica universalis, of which it would have formed a central part.
But it was almost two centuries after his death before his idea of a logical
calculus really sufficient for the kind of reasoning occurring in the exact
sciences was put into effect (in some form at least, if not the one Leibniz
had in mind) by Frege and Peano.
"It seems reasonable to suspect that it is this incomplete understanding of
the foundations which is responsible for the fact that mathematical logic
has up to now remained so far behind the high expectations of Peano and
others who (in accordance with Leibniz's claims) had hoped that it would
facilitate theoretical mathematics to the same extent as the decimal system
of numbers has facilitated numerical computations. For how can one expect to
solve mathematical problems systematically by mere analysis of the concepts
occurring if our analysis so far does not even suffice to set up the axioms?
But there is no need to give up hope. Leibniz did not in his writings about
the Characteristica universalis speak of a utopian project; if we are to
believe his words he had developed this calculus of reasoning to a large
extent, but was waiting with its publication till the seed could fall on
fertile ground. He went even so far as to estimate the time which would be
necessary for his calculus to be developed by a few select scientists to
such an extent "that humanity would have a new kind of an instrument
increasing the powers of reason far more than any optical instrument has
ever aided the power of vision." The time he names is five years, and he
claims that his method is not any more difficult to learn than the
mathematics or philosophy of his time. Furthermore, he said repeatedly that,
even in the rudimentary state to which he had developed the theory himself,
it was responsible for all his mathematical discoveries; which, one should
expect, even Poincare would acknowledge as a sufficient proof of its
I believe that what Leibniz could have had in mind is properly viewed, in
modern terms, as foundational thinking, in the sense that we have come to
know it and see it applied to effectively in the foundations of mathematics.
However, it was just too difficult - even for Leibniz - to be able to put
his vision into effect, until that vision had been put into dramatic and
systematic effect in the limited realm of f.o.m.
For all I know, a careful historical reading might show us that foundational
thinking cannot be what Leibniz had in mind.
But nevertheless, I propose that foundational thinking - in the sense that
we have come to know it through its dramatic applications to the foundations
of mathematics - is the appropriate modern form of what Leibniz might have
had min mind.
Some other comments on these quotations.
1. Note that Godel draws a distinction between two aspects of mathematical
logic. Its role as a branch of mathematics, and its role as the most
fundamental of all sciences.
I made essentially the same point in my parenthesized paragraph above, but
with different terminology that I think properly reflects the explosion of
activity since he wrote the quoted passages in 1944.
In 1944, there was not the kind of chasm that exists today between
mathematical logic and foundations of mathematics - and thus a chasm between
work that is called mathematical logic, and foundations of mathematics.
2. Concerning "For how can one expect to solve mathematical problems
systematically by mere analysis of the concepts occurring if our analysis so
far does not even suffice to set up the axioms?"
See my efforts in
Note that this work is suggestive of a *foundational process* which leads to
the setting up of axioms that at least interprets the axioms Godel is
talking about - the ZFC axioms.
It is not easy to state just what this foundational process is, but one
major aspect of it seems to be a recurring theme for me, and is discussed
explicitly in the above reference: simplicity investigations.
3. Concerning "Leibniz did not in his writings about the Characteristica
universalis speak of a utopian project; if we are to believe his words he
had developed this calculus of reasoning to a large extent, but was waiting
with its publication till the seed could fall on fertile ground."
I certainly do not want to claim, e.g., that, if only we developed
foundational thinking much further and much more systematically, then we
would quickly see how to prove the Riemann Hypothesis.
Foundational thinking, however, should greatly facilitate other kinds of
great advances of great general intellectual interest.
4. Concerning "Furthermore, he said repeatedly that, even in the rudimentary
state to which he had developed the theory himself, it was responsible for
all his mathematical discoveries; which, one should expect, even Poincare
would acknowledge as a sufficient proof of its fecundity."
I have certainly said many times that foundational thinking, with certain
tools and techniques, has been responsible for the preponderance of whatever
I have been able to do in f.o.m. I will try to spell out some specific tools
and techniques in later postings.
In the next postings on foundational thinking, I want to address the
following issues, not necessarily in this order.
1. What is foundational thinking? How does it differ from other kinds of
thinking? What is its relationship with mathematics and philosophy?
2. What are some examples of foundational thinking across a variety of
3. What, in detail, has been the great successes of foundational thinking in
4. What can we learn about foundational thinking from the f.o.m. example?
5. What are the main methods and tools of foundational thinking?
6. Will foundational thinking really advance our knowledge or perfect our
7. Will foundational thinking ever attain the level of depth and
effectiveness and power in connection with subjects outside mathematics?
8. Will the experience with foundational thinking in f.o.m. usably transfer
to other contexts?
9. What is foundational exposition?
10. What will a foundational exposition of mathematics look like?
11. What will a foundational exposition of subjects like statistics,
physics, economics, computer science, law, etc. look like?
12. What impact will foundational exposition have on systematic knowledge
and the creation of artificial systems?
13. What impact will foundational exposition have on education?
Harvey Friedman
More information about the FOM mailing list
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Ch. 11 Quiz
Ch.11 Quiz
This assessment is worth 100 points.
1. A simple discount note does not involve a bank discount.
(4 points)
2. A simple discount note results in a higher interest rate (effective) than a simple interest note.
(4 points)
3. The purchase price (or proceeds of a Treasury Bill) would be the value of the Treasury Bill plus the discount.
(4 points)
4. The calculation of the bank discount when discounting an interest-bearing note does use maturity value.
(4 points)
5. Proceeds from discounting an interest-bearing note is the principal less the bank discount.
(4 points)
6. The effective rate of a $25,000 non-interest-bearing simple discount 10 percent, 90 day note is:
(5 points)
7. When discounting an interest-bearing note the discount period represents:
(5 points)
8. A $9,000, 11 percent, 180-day note, dated March 18, is discounted on July 8. Assuming a 9 percent discount rate the bank discount is:
(5 points)
9. A $25,000, 15 percent, 80-day note, dated November 5, is discounted at National Bank on January 5. The discount period is:
(5 points)
10. Murray discounts a 120-day note for $20,000 at 11 percent. The proceeds would be: (Assume ordinary interest)
(5 points)
11. Shelley Corporation discounted a $7,000, 90-day note, dated June 18, at the Sunshine Bank on July 18 at a discount rate of 12 percent. (Assume the $7,000 is the maturity value) The amount of Bank
discount is:
(5 points)
12. Compute
(a) Bank discount; (b) Proceeds for the following simple discount (use ordinary interest); and (c) the effective interest rate to nearest hundredth percent.
(face) (disc. rate) (time)
(5 points)
(* 4000 character limit)
13. Calculate
Maturity value for this interest-bearing note using ordinary interest.
$48,000 82 days 12%.
(5 points)
(* 4000 character limit)
14. Mobilee Oil Company accepted a $10,000, 120-day note, dated March 3, at 8 1/2% to settle a past due account receivable. Mobilee Oil discounted the note to raise cash on May 10 at a discounted
rate of 9%. What proceeds did Mobilee Oil receive?
(5 points)
(* 4000 character limit)
15. On October 18, 2001 Blue Ridge Corporation accepted a $300,000 non-interest-bearing note from Long Corporation. What is the maturity value of the note?
(5 points)
(* 4000 character limit)
16. Wayne Night signed a $10,000 note at Lynn Bank that charges a 7% discount rate. If the loan is for 150 days, find:
A. Proceeds
B. Effective rate charges by the bank (to the nearest tenth percent).
(5 points)
(* 4000 character limit)
17. Compute bank discount, proceeds (use ordinary interest) and effective interest rate (to nearest hundredth). Do not round denominator in your calculation.
$12,000 13% 120 days
(5 points)
(* 4000 character limit)
18. Calculate maturity value for the interest bearing note using ordinary interest.
$36,000 92 days 14%.
(5 points)
(* 4000 character limit)
19. On May 12, Joy Co. accepted a $1,000, 60-day, 6 percent note from Abe Wills, granting a time extension on a past due account. Joy discounted the note at the bank at 9 percent on May 28. Calculate
Joy's proceeds.
(5 points)
(* 4000 character limit)
20. On October 15, Daniel Miller accepted a $5,000, 60-day, 8 percent note from Bill Boyer granting a time extension on a past due amount. Daniel discounted the note at Volve Bank at 9 percent on
Oct. 26. Calculate Daniel's proceeds.
(5 points)
(* 4000 character limit)
21. Molly Lenny bought a $10,000 13 week treasury bill at 13%. What is her effective rate? Round to nearest hundredth percent.
(5 points)
(* 4000 character limit)
Portions copyright ©2005 The McGraw-Hill Companies.
Any use is subject to the Terms of Use and Privacy Policy.
McGraw-Hill Higher Education is one of the many fine businesses of The McGraw-Hill Companies.
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MathGroup Archive: October 2007 [00896]
[Date Index] [Thread Index] [Author Index]
Why does Sum return 0 on this series?
• To: mathgroup at smc.vnet.net
• Subject: [mg82537] Why does Sum return 0 on this series?
• From: Adam Weyhaupt <aweyhau at siue.edu>
• Date: Wed, 24 Oct 2007 04:24:13 -0400 (EDT)
In Mathematica 6.0.1 on Mac OS X, Sum returns 0 on this series.
In[1]:= Sum[Log[n]^4/n^2, {n, 2, Infinity}]
Out[1]= 0
In[2]:= $Version
Out[2]= "6.0 for Mac OS X x86 (32-bit) (June 19, 2007)"
Of course that's not right, because the terms are all positive. I
don't expect Sum to always return a correct answer for any series
(that would be unreasonable), and NSum is the more appropriate
command to investigate this series numerically. But 0 seems like a
very strange answer. Can anyone provide any insight on why
Mathematica is returning 0 for the sum of a positive series?
Adam Weyhaupt
Department of Mathematics and Statistics
Southern Illinois University Edwardsville
aweyhau at siue.edu
Science Building 1316 (618) 650-2220
• Follow-Ups:
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Component X-over ?'s
01-06-2007 #1
Component X-over ?'s
How does a x-over manage to produce an ohm load of 4 ohms to an amp if there are two drivers connected to it?
What if you mix 2 or 4 ohm drivers on a x-over?
What if you connected two mids to the same x-over?
Or what if you didn't connect a tweeter, does that change the ohm load?
Do x-over's have a power limit?
Since X-over's just offer different frequency cutoff at different slopes why is it important to match drivers to the x-over's?
08 Evo X
Stock for now
Ref's: enzowho, rem, bdubs767, skydeaner
Re: Component X-over ?'s
I didn't write this, but it's probably the easiest to understand explaination I've seen;
The woofer has a lowpass in series with it, the tweeter has a highpass in series with it, and those 2 "assemblies" are wired in parallel. Now think about how the circuit behaves at various
frequencies, let's start at DC.
At DC the woofer's lowpass has an impedance of 0 and the woofer has an impedance of its Re (around 3.3ohm), since they're wired in series the woofer circuit has an impedance of 3.3ohm. At DC the
tweeter's highpass has an infinite impedance and the tweeter has an impedance of its Re, since they're wired in series, the tweeter circuit has an infinite impedance. Now as far as the amp is
concerned, you have a 3.3ohm impedance wired in parallel with an infinite impedance, and the result is 3.3ohm.
At very high frequencies you have the opposite, the tweeter's highpass has an impedance of 0 and the tweeter has an impedance around 4 or 8ohm depending on whether it's a 4 or 8ohm tweet. On the
other hand the woofer's lowpass has an infinite impedance, the woofer has something like a 10-20ohm impedance (it rises at high frequencies due to the inherent inductance), and so now the
woofer's circuit has an infinite impedance and the tweeter's circuit has a 4 or 8ohm impedance, wired in parallel you just end up with the tweeter.
At frequencies around the middle you get a mix, up until the crossover frequency the HP and LP filters will be very close to 0 and infinity respectively (or vice versa depending on which side of
the xover freq you're at). When you get close to the xover frequency, the filters start to switch, the one that was 0 starts to rise towards infinity, and the one that was very very high starts
to fall towards 0. At the crossover frequency, if you have say 4th order LR filters, each filter's impedance will be exactly equal to the impedance of the speaker that it's controlling. So the
woofer's lowpass filter will be equal to 4ohm, and the tweeter's highpass filter will be equal to the tweeter's impedance. Since the filter is in series with the speaker, the woofer's effective
impedance will be double, and the tweeter's effective impedance will be double. When you wire these two in parallel, the result will be right around the nominal impedance of either of them.
If you have a 4ohm woofer and a 4ohm tweeter, the final impedance should be right around 4ohm throughout the entire frequency spectrum.
What if you mix 2 or 4 ohm drivers on a x-over?
If you place a 2ohm speaker on a passive xover designed for a 4ohm load, then the xover frequency would be cut in half (this is not desireable).
If the xover was intentionally designed to be mated with a 2ohm midwoofer and a 4ohm tweeter, for example, then when for frequencies within the mids passband the amplifier would see a 2ohm load
and for frequencies within the tweeters passband it would see a 4ohm load.
What if you connected two mids to the same x-over?
If you connected them in parallel, the xover frequency would be cut in half. If you wired them in series, the xover frequency would be doubled.
Or what if you didn't connect a tweeter, does that change the ohm load?
Bad idea.
Depending on the design of the crossover, leaving one of the circuits open could result in the amplifier being presented with a dead short, consequently risking damage to the amplifier.
Do x-over's have a power limit?
There will be a limit where the components within the crossover can not handle the power being passed through them. But in normal circumstances, it isn't something the consumer should really
"worry" about as it is likely much higher than the amount of power being applied to them.
Since X-over's just offer different frequency cutoff at different slopes why is it important to match drivers to the x-over's?
Well, for one; see above. If you connect an improper impedance load for the xovers design, you will affect the xover frequency and as such the resulting response will not be desireable.
Likewise, all speakers will perform optimally with certain xover frequencies and slopes specific to each individual speaker. Mixing speakers with xover, even if impedances are taken into
consideration, may not result in optimal performance or response from the speakers. Also, if you place a tweeter that requires a relatively high xover frequency and slope on a passive xover that
has a lower-than-recommended slope or frequency; you risk damaging the tweeter.
Lastly; Xovers can do MUCH more than just offer differing slopes and xover frequencies. Many have built in impedance compensation (zobel networks), passive EQ circuits, notch filters, etc etc.
All of which were designed to provide optimal performance from one specific speaker, and not necessarily any speaker you choose to place on the passive. Again, this is all about acheiving optimal
performance from the speaker; and placing just "any" speaker on a specifically designed passive xover very well may not acheive this.
Re: Component X-over ?'s
wow..Great response. Thank you for taking the time to answer my questions.
So for people running 2 or 3 mids and one tweeter in their cars how are they acheiving this? Actively?
08 Evo X
Stock for now
Ref's: enzowho, rem, bdubs767, skydeaner
Re: Component X-over ?'s
I wrote a good bit on this a while back, a search would probably locate it.
As for running multiple mids, it can be done a few different ways. Active is probably the easiest. Custom designed passives are another way to go about it. Back in the late 80's/early 90's it was
fairly common for SQ comp vehicles to run huge numbers of drivers on a single 2 channel amp. Huge and hugely complex passive networks were needed to get this to work right and sound good.
Squeak minor point on your post about the effects of using the wrong impedance driver on a passive. Your answer was correct for a 1st order filter but higher orders behave really strangely with
the wrong load on them. One half of the filter shifts one way and the other half shifts the other. The result is a stepped filter and a probably ****** response curve.
Artificial Intelligence is no match for Natural Stupidity.
Never underestimate the power of stupid people in large numbers.
Life's tough...it's even tougher when you're stupid.
Re: Component X-over ?'s
I was trying to keep it relatively simple
Re: Component X-over ?'s
Realtively simple would be "It won't work right."
Artificial Intelligence is no match for Natural Stupidity.
Never underestimate the power of stupid people in large numbers.
Life's tough...it's even tougher when you're stupid.
01-06-2007 #2
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To: Reiny- Math
Posted by Anonymous on Wednesday, May 22, 2013 at 11:28pm.
Thank you so much for helping me but for #2 c.) is it yes or a no, though i did the rest of my questions but im stuck on this one so can you please help and explain a lot cause i have a test on this
soon and i want to understand it, thanks :)
Serena can rent a video game for $3.49 per day. She can buy the game for $49.95. After how many days does it become cheaper for Serena to buy the game?
a.) use an inequality to reprsent the situation
b.) use the ineqality to solve the problem
c.) if the game takes serena 25h to solve and she plays 1.5h a day, should she rent or buy the game? verify your solution.
Thank you
• To: Reiny- Math - Anonymous, Thursday, May 23, 2013 at 12:00am
I tried doing the question: are these correct:
for a.) 3.49x< or equal to 49.95
b.) 49.95/3.49= 14.3= 14 days
and im stuck on c.)
• To: Reiny- Math - Anonymous, Thursday, May 23, 2013 at 12:10am
oh for c.) i got: she should buy the game because in 14 days she will have only completed 21 hours of the game: 1.5 x 14= 21
25/1.5 = 16 days to complete the game, therefore she should buy it
but can you check these to see if im right and thank you so much for helping me, it means a lot and i have finally understood how to do it thank you:):):):)
• To: Reiny- Math - Reiny, Thursday, May 23, 2013 at 1:07am
Since you did not sign with a name and used "anonymous" instead, it is hard to go back and find which post contains #2 c)
As to the above ....
a) and b) are correct
if you get a decimal, like in x < 49.35/3.49 < 14.3
actually try the whole number above and below to see which one works
c) Serena will need 25/1.5 or 16 2/3 days to complete the game, but again, we need whole numbers .
So she will need to rent the game for 17 days, not 16
the cost for 17 days would be 17(3.49) = 59.33 , so yes, she should buy it for only 49.95
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math and litetature - Just wanted to say thank you to all of you for helping us...
thank you,write teacher for helping my letter - thank you very much.
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Equation of circle
March 17th 2008, 04:04 PM
Equation of circle
Find the equation of the circle which passes through (-1, 4)and with center at (4,-8)
I know this is an easy question but I just cant find the formula for how to find the equation. Any help appreciated.
March 17th 2008, 04:08 PM
I think you'd find the distance from the center to that point. That's the radius.
March 17th 2008, 04:31 PM
The standard form for the equation of a circle is:
$(x - h)^2 + (y - k)^2 = r^2$
Where the point $(h,k)$ is the center of the circle.
We are given that the center is found at (4,-8) so we have our equation so far:
$(x - 4)^2 + (y + 8)^2 = r^2$
Now we just need to find the radius. We have been given one point on the circle, so to find the radius we simply need to find the distance from the center of the circle to that point. Recall the
distance formula:
$d = \sqrt {(x_2-x_1)^2+(y_2-y_1)^2}$
Plug your values in to find the radius, then simply enter that into the previous equation to get your final answer. :)
March 17th 2008, 04:36 PM
Ah. The answer is $(x-4)^2 + (y+8)^2 = 169$ from what you said about the distance from those 2 points.
March 17th 2008, 04:45 PM
That is correct.
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Aluminized Steel Plate
If you want to drastically oversimplify things like that, then yes as a matter of fact with my current setup I have preheated my 1/2" steel to over 700F in 10 minutes. But it doesn't work like that.
For one most people are using 1/2" steel with nearly twice the thermal mass of your plate, meaning it will have nearly twice the preheat time. Second, every oven is different. I know you have tried
multiple ovens with good results, but if the standard gas ovens you have used it with are out performing your commercial quality convection oven I am going to guess they just so happened to be pretty
powerful ovens and lead to a fast preheat. I would guarantee some of the anemic ovens many users here have would not give the same results.
Understand I'd love to be wrong here, I'd love for there to be an even better material out there. There is just no theoretical reason for this material to be better then standard steel, and there I
have not personally seen anything posted in this thread that changes that. That all said I am curious to see and hear about anything you can do to change my mind because I truly would love to be
wrong about this one.
Just one question. Does anybody else preheat only to the cut off in the oven and obtain same temperaure?
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Topic: Assigning different values for variable issue
Replies: 1 Last Post: Feb 1, 2013 9:43 AM
Re: Assigning different values for variable issue
Posted: Feb 1, 2013 9:43 AM
"Matthew " <mdforman91@hotmail.com> wrote in message
> So what I'm trying to do is simulate different damping ratios (labeled
> "C") and simply use those in a bunch of equations for outputting a plot of
> the displacement. However I'm having matrix agreement issues and I can't
> figure out why. My code is:
> t=0:0.1:5;
> C=[.01 .2 .6 .1 .4 .8];
> Wn=2;
> Wd=Wn*sqrt(1-C.^2);
> A=sqrt(((C.*Wn*1).^2 + Wd.^2)./(Wd.^2));
> phi=atan(Wd./(C.*Wn*1));
> x=A.*exp(-C.*Wn*t).*sin(Wd.*t + phi);
> plot(x,t)
> I've found that the issue takes place in my "x" equation between Wn*t and
> I can't figure out what is wrong. Excuse all the dot assignments, as I'm
> not sure if all them were needed or not.
.* is elementwise multiplication. The two quantities being multiplied must
be EXACTLY the same size (in which case the result is that same size) or one
must be scalar (in which case the result is the size of the other.)
* is matrix multiplication. The number of columns in the first quantity
being multiplied must be equal to the number of rows in the second (in which
case the result has the same number of rows as the first and the same number
of columns as the second), or one must be scalar (in which case the result
is the size of the other.)
Since Wn is scalar the multiplication C.*Wn is valid and results in a vector
the same size as C. Then (C.*Wn)*t is attempting to perform matrix
multiplication on two vectors that do NOT satisfy the requirements, and
that's the cause of the problem. I'm guessing you want to perform
elementwise multiplication there, but C and t are not the same size so that
won't work. You may want to change your t to something like:
t = 0:1:5;
and use elementwise multiplication.
Steve Lord
To contact Technical Support use the Contact Us link on
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Covering a Cube with a Square
up vote 11 down vote favorite
Suppose you are given a single unit square, and you would like to completely cover the surface of a cube by cutting up the square and pasting it onto the cube's surface.
Q1. What is the largest cube that can be covered by a $1 \times 1$ square when cut into at most $k$ pieces?
The case $k=1$ has been studied, probably earlier than this reference: "Problem 10716: A cubical gift," American Mathematical Monthly, 108(1):81-82, January 2001, solution by Catalano-Johnson, Loeb,
(This was discussed in an MSE Question.) The depicted solution results in a cube edge length of $1/(2\sqrt{2}) \approx 0.35$.
As $k \to \infty$, there should be no wasted overlaps in the covering of the 6 faces, and so the largest cube covered will have edge length $1/\sqrt{6} \approx 0.41$. What partition of the square
leads to this optimal cover?
Q2. For which value of $k$ is this optimal reached?
I have not found literature on this problem for $k>1$, but it seems likely it has been explored. Thanks for any pointers!
mg.metric-geometry geometry recreational-mathematics reference-request
I wonder if anyone in the packaging industry has the answer. – Steven Gubkin May 4 '12 at 0:37
I believe there was a "Mathematical Games" column on dissections which had a Greek cross rearranged into a square. Perhaps that article also mentioned this problem? Gerhard "Testing Your Martin
Gardner Fu" Paseman, 2012.05.04 – Gerhard Paseman May 4 '12 at 15:51
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3 Answers
active oldest votes
You can cut a $\sqrt{6}\times\sqrt{6}$ square into 24 pieces that then cover the $1\times1\times1$ cube. Two triangles from the figure below plus one parallelogram make up one $1\
times1$square. parts of pieces sticking out to the left can obviously fit back in the right, so 18 pieces, plus 6 parts sticking out equals 24. You can improve on this by stitching
pieces across the cube edge to make one bent piece and by stitching some of the parallelograms back to the triangles.
[Added by O'Rourke:] Just to make Yoav's construction more explicit, here is how two triangles and a parallelogram fit together to form a $1 \times 1$ square:
up vote 17
down vote [Added by Kallus:] Here's an illustration of a construction similar to Fedja's construction but with only five pieces. The first figure is the $\sqrt{6}\times\sqrt{6}$ square. The second
accepted is the $2\times3$ rectangle, which we fold into a cube by taking away the two yellow squares, folding the remainder, and adding the squares as the two missing faces.
[Added by O'Rourke:]
1 Brilliant!! :-) – Joseph O'Rourke May 3 '12 at 23:51
5 Actually, any two polygons of the same area are equidecomposable and the surface of the cube can be unfolded into a polygon, so the result is nice but not terribly surprising. Of
course, the question about the minimal number of pieces remains. – fedja May 4 '12 at 0:15
2 Assuming that "pieces" mean "connected polygonal pieces", we can take a 3 by 2 rectangle, cut it into a T-shape and two unit squares and then use the standard "sliding cut" to turn it
into a square, giving the total of 6 pieces to cover the unit cube. Can we do better? – fedja May 4 '12 at 0:34
2 OK. Posted to soon earlier. The change from the T tetromino to the S tetromino actually allows going down to five pieces instead of six. – Yoav Kallus May 4 '12 at 4:21
1 Wow! $\mbox{}$ – Joseph O'Rourke May 4 '12 at 10:18
show 4 more comments
Sorry, this is an answer to an other question. (I did not read the question carefully.)
Question: For which $k$, $k$ squares can tile the surface of cube.
Answer: $k=6\cdot(n^2+m^2)$.
up vote 9 down Here is a tiling with $k=30$, $n=1$ and $m=2$.
It is obvious if the tiling is vertex-to-vertex.
If the tiling is not vertex-to-vertex, you get a closed geodesic formed by overlaping sides. Then you can shift squares on one side of the geodesic to make the tiling "more
vertex-to-vertex". Repeating this operation you can make the tiling to be vertex-to-vertex.
@Anton: Did you intend $n \ge 1 , m \ge 1$, or, say, $n \ge 1 , m \ge 0$ ? – Joseph O'Rourke May 3 '12 at 18:03
Yes, $k$ has to be positive; so $n\ge 1$ and $m\ge 0$. – Anton Petrunin May 3 '12 at 18:33
Awesome ! – Steven Gubkin May 3 '12 at 19:05
@Anton: Sorry to be slow :-/, but could you describe a partition of the square into 6 pieces that exactly cover the cube? – Joseph O'Rourke May 3 '12 at 19:21
@Joseph: the partition into faces (the cube has 6 faces). – Anton Petrunin May 3 '12 at 19:35
show 3 more comments
No promises that these are optimal, but here are some lower bounds:
up vote 2 down vote With $k=2$, side length $3/8=0.375$ (with one piece flipped over), and with $k=3$, side length $2/5=0.4$:
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Not the answer you're looking for? Browse other questions tagged mg.metric-geometry geometry recreational-mathematics reference-request or ask your own question.
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What exactly is centrifugal force
They extend the applicability of Newton's 1nd & 2nd to non-inertial frames.
No, they don't. As vector equations, Newton's laws hold in any coordinate system or frame. It's always the case that in the absence of external forces, the velocity vector [itex]\stackrel{\
rightarrow}{U}[/itex] satisfies
[itex]\frac{\stackrel{\rightarrow}{dU}}{dt} = 0[/itex]
It's always the case that in presence of a (real, non-fictitious) force [itex]\stackrel{\rightarrow}{F}[/itex], the velocity satisfies
[itex]m \frac{\stackrel{\rightarrow}{dU}}{dt} = \stackrel{\rightarrow}{F}[/itex]
It's always the case that if one object exerts a (real, non-fictitious) force [itex]\stackrel{\rightarrow}{F}[/itex] on another, then the second exerts a force [itex]-\stackrel{\rightarrow}{F}[/itex]
on the first.
Newton's laws, as vector equations are not changed at all by using noninertial or curvilinear coordinates. Fictitious forces are not needed to extend Newton's laws to noninertial frames. What's
needed is a more sophisticated notion of what
means when the basis vectors themselves are nonconstant.
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From HaskellWiki
A perpetual Haskell newbie. I like this semi-one-liner:
-- inifinte folding idea due to Richard Bird
-- double staged production idea due to Melissa O'Neill
-- tree folding idea Dave Bayer / simplified formulation Will Ness
primes = 2 : g (fix g)
g xs = 3 : gaps 5 (foldi (\(c:cs) -> (c:) . union cs) []
[[x*x, x*x+2*x..] | x <- xs])
fix g = xs where xs = g xs -- global defn to avoid space leak
gaps k s@(c:t) -- == minus [k,k+2..] (c:t), k<=c,
| k < c = k : gaps (k+2) s -- fused to avoid a space leak
| True = gaps (k+2) t
foldi is on Tree-like folds page. union and more at Prime numbers.
The constructive definition of primes is the Sieve of Eratosthenes:
$\textstyle\mathbb{S} = \mathbb{N}_{2} \setminus \bigcup_{p\in \mathbb{S}} \{p\,q:q \in \mathbb{N}_{p}\}$
using standard definition
$\textstyle\mathbb{N}_{k} = \{ n \in \mathbb{N} : n \geq k \}$ . . . or, $\textstyle\mathbb{N}_{k} = \{k\} \bigcup \mathbb{N}_{k+1}$ :):) .
Trial division sieve is:
$\textstyle\mathbb{T} = \{n \in \mathbb{N}_{2}: (\forall p \in \mathbb{T})(2\leq p\leq \sqrt{n}\, \Rightarrow eg{(p \mid n)})\}$
If you're put off by self-referentiality, just replace $\mathbb{S}$ or $\mathbb{T}$ on the right-hand side of equations with $\mathbb{N}_{2}$, but even ancient Greeks knew better.
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VTU DSP Algorithms and Architecture Previous question papers 7th sem
VTU DSP Algorithms and Architecture Previous question papers 7th sem EC751 ECE
Tags : VTU Question Papers, dsp algorithms and architecture vtu, dsp algorithms and architecture vtu question papers.
VTU DSP Algorithms and Architecture Previous question papers 7th sem
VTU DSP Algorithms and Architecture Previous question papers:DSP Algorithms and Architecture can download the following Question papers.we have given DSP Algorithms and Architecture Previous question
papers is an important subject in ECE Department VTU Question papers is given by student Reference.All previous year Question papers in PDF Format. ECE 7th Semester DSP Algorithms and Architecture
Question papers is given by Kin India Students can download.EC751 ECE Question papers may/june and nov/dec exam is given below link.
VTU DSP Algorithms and Architecture question papers
University : Visvesvaraya Technological University
Branch : ECE
Semester : 7
Subject : DSP Algorithms and Architecture
Subject code :EC751
Scheme : 2010 and 2006
dsp algorithms and architecture vtu question papers dec09/jan10
If you have any doubts or need any material, comment below.
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Computing the constant D-module on an intersection
up vote 5 down vote favorite
Let $M$ be a smooth variety, say over the complex numbers, and let $i:W \hookrightarrow M, j: Z \hookrightarrow M$ be smooth subvarieties. Let $i_{+},j_{+}$ denote (derived) pushforward of D-modules,
$i^{*},j^{*}$ (derived) pullback.
Question, in general: How do I compute the (complex) of $D$-modules $i_{+}\mathcal{O}_{W} \otimes^{L} j_{+} \mathcal{O}_{Z} \simeq i_{+}i^{*} \mathcal{O}_{M} \otimes^{L} j_{+}j^{*} \mathcal{O}_{M}$?
If $X=W \cap Z$ is smooth and of the expected codimension, then base change and the projection formula imply that $i_{+}\mathcal{O}_{W} \otimes^{L} j_{+} \mathcal{O}_{Z} \simeq i_{+}i^{*} \mathcal{O}
_{M} \otimes^{L} j_{+}j^{*} \mathcal{O}_{M} \simeq h_{+}h^{*}\mathcal{O}_{M} \simeq h_{+} \mathcal{O}_{X}$, where $h: X \rightarrow M$ is the inclusion. Thus we get the constant $D$-module supported
along $X$.
Actually, I'm most interested in the case when $X$ is not necessarily smooth or of the expected codimension. Then the answer for the corresponding question for $\mathcal{O}$-modules is $i_{*}\mathcal
{O}_{W} \otimes^{L} j_{*}\mathcal{O}_{Z}$, which is the structure sheaf of the derived intersection of $W$ and $Z$, and its cohomology sheaves are just $Tor_{k}(i_{*}\mathcal{O}_{W}, j_{*}\mathcal{O}
_{Z})$, and (in theory and often in practice) I know how to compute these by writing down a locally free resolution of $i_{*}\mathcal{O}_{W}$ as an $\mathcal{O}_{M}$-module.
In theory, I also know that I can take a resolution of $i_{+}\mathcal{O}_{W}$ by $D$-modules that are locally free as $O$-modules. But in practice I don't know how to do this.
Question, more specific: Suppose I have a locally free resolution of the $\mathcal{O}_{M}$-module $i_{*}\mathcal{O}_{W}$. Can I use this to build a $D$-module resolution that is locally free over $\
Question, very specific: Let $W, M=T^{*}W$, and $i: W \hookrightarrow T^{*}W$ is the zero section. Then I have the standard Koszul resolution of $i_{*}\mathcal{O}_{W}$. Is there a 'Koszul resolution'
of the $D$-module $i_{+}\mathcal{O}_{W}$? More generally, you could replace $T^{*}W$ with any vector bundle.
d-modules ag.algebraic-geometry
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Browse other questions tagged d-modules ag.algebraic-geometry or ask your own question.
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How many lines per centimeter does a grating have if the... (1 Answer) | Transtutors
Light and diffraction
How many lines per centimeter does a grating have if the third-order occurs at an 17.0 degree angle for 630 nm light?
Posted On: Mar 17 2012 04:49 PM
Tags: Science/Math, Physics, Classical Physics, College
1 Approved Answer
The formula for diffraction grating maxima is
dsinθ = mλ
Solving for d we get
d = mλ/sinθ = (3)(620 X 10^-9)/(sin 17) = 6.36 X 10^-6 m/line which is 6.36 X 10^-4 cm/line
Since we are asked for lines per cm, then we need to take the inverse of d
1/ 6.36 X 10^-4 cm/line = 1.57 X 10^3 lines/cm
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Configuration space of little disks inside a big disk
up vote 20 down vote favorite
The space of configurations of $k$ distinct points in the plane $$F(\mathbb{R}^2,k)=\lbrace(z_1,\ldots , z_k)\mid z_i\in \mathbb{R}^2, i\neq j\implies z_i\neq z_j\rbrace$$ is a well-studied object
from several points of view. Paths in this space correspond to motions of a set of point particles moving around avoiding collisions, and its fundamental group is the pure braid group. It is not hard
to prove that this space is homotopy equivalent to the configuration space of the unit disk $$F(D^2,k)=\lbrace(z_1,\ldots , z_k)\mid z_i\in D^2, i\neq j\implies z_i\neq z_j\rbrace$$
In real life however, particles, or any other objects which move around in some bounded domain without occupying the same space, have a positive radius, and so would be more realistically modelled by
disks rather than points. This motivates the study of the spaces $$F(D^2,k;r)= \lbrace(z_1,\ldots , z_k)\mid z_i\in D^2, i\neq j\implies |z_i - z_j|>r\rbrace$$ where $r>0$. The homotopy type of this
space is a function of $r$ and $k$. Fixing $k$ and varying $r$ gives a spectrum (is this the right word?) of homotopy types between $F(\mathbb{R}^2,k)$ and the empty space. It seems like an
interesting (and difficult) problem to study the homotopy invariants as functions of $r$. For instance, for $k=3$ what is $\beta_1(r)$, the first Betti number of $F(D^2,k;r)$?
Question: Have these spaces and their homotopy invariants been studied before? If so, where?
Of course one can also ask the same question with disks of arbitrary dimensions.
at.algebraic-topology configuration-space reference-request
1 I wrote a detailed answer below including an example with hard disks in a square, but we also know some things about hard disks in a disk. Even though the boundary is smooth, things still get
complicated pretty quickly. I have pictures of all the "critical configurations" we have found for disks in a disk, up to about seven disks, and I can email them on request. – Matthew Kahle Mar 25
'11 at 21:32
Persi Diaconis mentioned a similar problem viewed from a slightly different perspective in his paper "The Markov chain Monte-Carlo revolution" – Simon Lyons Mar 25 '11 at 21:54
1 My answer was going to be "Ask Matt Kahle," but since he's already answered, I'll just link to what I wrote about this problem on my blog: in particular, its conceptual affinity with the "15
puzzle." quomodocumque.wordpress.com/2009/04/04/… – JSE Mar 26 '11 at 20:41
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3 Answers
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I have a number of results on hard disks in various types of regions, and preprints are in progress. The terminology "hard spheres" (or "hard disks" in dimension 2) comes from
statistical mechanics, and I believe Fred Cohen is following my lead on this. (See for example the hard disks section of Persi Diaconis' survey article.)
--- With Gunnar Carlsson and Jackson Gorham, we did numerical experiments and computed the number of path components for 5 disks in a box, as the radius varies over all possible values.
This is quite a complicated story already, as the number is not monotone or even unimodal in the radius. (This preprint is almost done, a rough copy is available on request.)
--- With Yuliy Baryshnikov and Peter Bubenik we develop a general Morse-theoretic framework and proved that in a square, if $r < 1/2n$ then the configuration space of $n$ disks of
radius $r$ is homotopy equivalent to $n$ points in the plane. On the other hand this is tight: if $r> 1/2n$ then the natural inclusion map is not a homotopy equivalence. (Again this
preprint is getting close to be being posted to the arXiv...) There is a much more general statement here.
--- I also have some results with Bob MacPherson about hard disks in a square and also in (the easier case of) an infinite strip. We have been talking to Fred Cohen about this a bit
lately, who believes there may be connections to more classical configuration spaces.
I am slightly self-conscious about claiming results here, without first having posted the preprints to the arXiv, but I just wanted to state that there are a number of things known now,
and I am working hard to get everything written up in a timely fashion. In the meantime I have some slides up from a talk I recently gave at UPenn.
Here is a concrete example, since that might be more satisfying.
It turns out for $3$ disks in a unit square:
up vote 23
down vote For $0.25433 < r$, the configuration space is empty,
for $0.25000 < r < 0.25433$ it is homotopy equivalent to $24$ points,
for $ 0.20711 < r < 0.25000$ it is homotopy equivalent to $2$ circles,
for $ 0.16667 < r < 0.20711 $ it is homotopy equivalent to a wedge of $13$ circles, and
for $ r < 0.16667$ it is homotopy equivalent to the configuration space of $3$ points in the plane.
For $4$ disks in a square it looks like the topology changes $9$ or $10$ times, and for $5$ disks it looks like the topology might change $25$-$30$ times or more. The general idea is
that certain types of "jammed" configurations act like critical points of a Morse function, and mark the only places where the topology can chance.
Update: two preprints have been posted to the arXiv.
Min-type Morse theory for configuration spaces of hard spheres (w/ Baryshnikov and Bubenik):
Computational topology for configuration spaces of hard disks (w/ Carlsson, Gorham, and Mason): http://arxiv.org/abs/1108.5719
1 I think most of your inqualities are backwards? It's clear what you mean. – Theo Johnson-Freyd Mar 26 '11 at 5:18
Thankyou for your nice answer and for linking to the slides. Now I remember hearing Baryshnikov talk about this (my visual memory was sparked by your picture of "Kahle's corner"). So
is it correct to say: (1) that these jammed configurations can't occur for disks in a disk or squares in a square; (2) that the models in (1) are equivalent, and harder than disks in
a square? – Mark Grant Mar 26 '11 at 8:05
Nice answer. I think that squares in squares is easier than discs in a square, I'd very much like to know where discs in discs lies, is it harder than squares in squares (I'd imagine
so). Is it only easier than discs in squares because it has the full orthogonal symmetry, rather than the dihedral symmetry? Finally, a fun application is to origami. Circle packings
in squares give optimum folds of animals with a fixed number of 'legs'. See Lang's Origami Design Secrets. – James Griffin Mar 26 '11 at 12:52
@ Theo: Thanks, fixed. – Matthew Kahle Mar 26 '11 at 13:21
@ Mark, James: I think squares in a region is generally easier than disks in a region, at least if we assume that the sides stay parallel to the axes. From a lot of experimenting
with small examples I believe that disks in a disk looks slightly easier than disks in a square at first, but that this is a little misleading --- it only takes a few more disks for
the complications to begin. For example: with 3 disks in a disk, the topology only changes twice. But by the time you get up to 7 disks in a disk, it looks like the topology changes
at least 18 times as the radius varies. – Matthew Kahle Mar 26 '11 at 13:26
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Yes, there has been a lot of work by Fred Cohen (University of Rochester, currently at IAS) on the subject. He has been giving talks about this (he calls it the hard sphere model), but I
can't seem to find a relevant paper/preprint. Perhaps you can contact him directly. EDIT I have somehow managed to confuse Cohen's work and Matt Kahle's (if you know them both, you know
up vote 6 how impressive a feat that is). Matt's answer is the true received wisdom.
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I would very much appreciate a good answer to this question, perhaps a follow up to Igor's answer as this is something that I have thought about before, but have not come across in
the literature.
I quickly (perhaps too quickly) abandoned the discs model in favour of the little cubes model, or perhaps I should say the hard cubes model.
For hard 2-cubes and k=3 I think the homology groups are:
up vote 5 down for r > 1/2: clearly 0
for 1/2 >= r > 1/3: H_0 = Z^6, H_1=Z^6 and H_i=0 for i>1
the three squares are effectively arranged in a circle which can be rotated, the order (and not just the cyclic order!) parametrises the 6 connected components.
for 1/3 >= r we get the usual configuration space: H_0 = Z, H_1 = Z^3, H_2 = Z^2 and H_i=0 for i>2.
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