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# Kaplan-Meier Curve using R

Published:

Kaplan-Meier curve shows what the probability of an event (for example, survival) at a certain time interval. The log-rank test compares the survival curves of two or more groups. With a small subset of patients, the Kaplan-Meier estimates can be misleading and should be interpreted with caution.

KM Analysis using R The packages used for the analysis are survival and survminer. Use install.packages( ) to install these libraries just in case if they are not pre installed in your R workspace.

library(survival)
library(survminer)
library(dplyr)

Examine the dataset (Vital status: 1 for dead; 0 for alive)

ID      p16 		Status	Days
GHN-11  unknown		1  	803
GHN-15  unknown		1  	775
GHN-20  unknown		1  	150
GHN-21  unknown		0 	2036
GHN-24  unknown		1  	718
GHN-25	negative	1  	598
HN-39	positive	0	1232

Create a survival object, usually used as a response variable in a model formula.

surv_obj = survival::Surv(time=data$Days, event = data$Status)

Wrapper arround the standard survfit() function to create survival curves

fit = survminer::surv_fit(surv_obj ~ p16, data = data)

You can replace the above two steps with

fit = surv_fit(Surv(Days, Status) ~ p16, data = data)

Plot the KM curve. With pval = TRUE argument, it plots the p-value of a log rank test, which will help us to get an idea if the groups are significantly different or not.

png(filename = paste(base.dir, "KM_Plot.png", sep="/"),width = 1300, height = 1300, res=200)
ggsurvplot(fit_1, pval=TRUE, risk.table=TRUE)
dev.off()

KM plot The lines represent survival curves of the 3 groups; HPV-status: positive [N=23], negative [N=5] and unknown [N=7]. A vertical drop in the curves indicates an event (eg. death). For HPV-positive: 5 (21.7%); HPV-negative: 4 (80%) and HPV-unknown: 5 (71.4%) events.

The lengths of the horizontal lines along the X-axis of serial times represent the survival duration for that interval. The vertical tick mark on the curves means that a patient was censored at this time; a patient has not (yet) experienced the event of interest, such as death, within the study time period. If many patients were censored in a given group(s), one must question how the study was carried out or how the type of treatment affected the patients. This stresses the importance of showing censored patients as tick marks in survival curves.

Risk table At time zero, the survival probability is 1.0 (or 100% of the participants are alive). At time 0, all 35 are alive or at risk and after 1000 days, there are 21 participants alive or at risk; after 3000 days, 3 participants are alive or at risk.

format: plain


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1. ## arc length problem

This is actually an arc length problem for the polar equation r=1+sin(x) (x being theta) After some work you get:
L=integral from 0 to 2 times pi (2+2*sin(x))^(1/2) dx. It gets reduced to

L=2^(1/2) times integral from 0 to 2 times pi (cos(x)/(1-sin(x))^1/2) dx

But if u=1-sin(x) don't the limits of integration both equal 1?

The book says the answer is L=8 The figure is a cardioid.

2. ## Re: arc length problem

Instead of simplifying $\sqrt{\cos^2{x}}=\cos{x}$, you should have used $\sqrt{\cos^2{x}}=|\cos{x}|,$ and then the integral splits into three parts: integrating from 0 to $\frac{\pi}{2}$, from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, and from $\frac{3\pi}{2}$ to $2\pi$. The answer 8 is correct.

- Hollywood

3. ## Re: arc length problem

To make it a little more clear what you are saying:

$L = \int_0^{2\pi} \sqrt{2+2\sin \theta}d\theta = \sqrt{2} \int_0^{2\pi} \left| \cos \theta \right| \dfrac{1}{\sqrt{1-\sin \theta}} d\theta$

Note the absolute value around $\cos \theta$ since $\sqrt{\cos^2 \theta} \ge 0$ for all $\theta$.

Since $\cos \theta \ge 0$ on $\left[0,\dfrac{\pi}{2} \right]\cup \left[\dfrac{3\pi}{2}, 2\pi \right]$ and $\cos \theta \le 0$ on $\left[ \dfrac{\pi}{2},\dfrac{3\pi}{2} \right]$, you need to break up the integral:

$L & = \sqrt{2}\left[ \int_0^{\tfrac{\pi}{2}} \dfrac{\cos \theta}{\sqrt{1-\sin \theta}}d\theta - \int_{\tfrac{\pi}{2}}^{\tfrac{3\pi}{2}} \dfrac{\cos \theta}{\sqrt{1-\sin \theta}}d\theta + \int_{\tfrac{3\pi}{2}}^{2\pi} \dfrac{\cos \theta}{\sqrt{1-\sin \theta}}d\theta \right]$

Using the substitution $u = 1-\sin \theta, du = -\cos \theta d\theta$, you get:

\begin{align*}L & = \sqrt{2}\left[ -\int_1^0 u^{-1/2}du + \int_0^2 u^{-1/2}du - \int_2^1 u^{-1/2}du \right] \\ & = \sqrt{2}\left[ \int_0^1 u^{-1/2}du + \int_0^2 u^{-1/2}du + \int_1^2 u^{-1/2}du \right] \\ & = 2\sqrt{2}\int_0^2 u^{-1/2}du \\ & = 4\sqrt{2}(\sqrt{2}-0) = 8\end{align*}

Edit: Someone else beat me to it.

4. ## Re: arc length problem

Wow ! thanks to both of you. I never considered that.


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The quadratrix or trisectrix of Hippias (also quadratrix of Dinostratus) is a curve, which is created by a uniform motion. It is one of the oldest examples for a kinematic curve, that is a curve created through motion. Its discovery is attributed to the Greek sophist Hippias of Elis, who used it around 420 BC in an attempt to solve the angle trisection problem (hence trisectrix). Later around 350 BC Dinostratus used it in an attempt to solve the problem of squaring the circle (hence quadratrix).

## Definition

Consider a square ABCD with an inscribed quarter circle centered in A, such that the side of the square is the circle's radius. Let E be a point that travels with a constant angular velocity on the quarter circle arc from D to B. In addition the point F travels with a constant velocity from D to A on the line segment AD, in such a way that E and F start at the same time at D and arrive at the same time in B and A. Now the quadratrix is defined as the locus of the intersection of the parallel to AB through F and the line segment AE.[1][2]

If one places such a square ABCD with side length a in a (cartesian) coordinate system with the side AB on the x-axis and vertex A in the origin, then the quadratix is described by a planar curve ${\displaystyle \gamma :(0,{\tfrac {\pi }{2}}]\rightarrow \mathbb {R} ^{2}}$ with:

${\displaystyle \gamma (t)={\begin{pmatrix}x(t)\\y(t)\end{pmatrix}}={\begin{pmatrix}{\frac {2a}{\pi }}t\cot(t)\\{\frac {2a}{\pi }}t\end{pmatrix}}}$

This description can also be used to give an analytical rather than a geometric definition of the quadratrix and to extend it beyond the ${\displaystyle (0,{\tfrac {\pi }{2}}]}$ interval. It does however remain undefined at the singularities of ${\displaystyle \cot(t)}$ except for the case of ${\displaystyle t=0}$, where due to ${\displaystyle \lim _{t\to 0}t\cot(t)=1}$ the singularity is removable and hence yields a continuous planar curve on the interval ${\displaystyle (-\pi ,\pi )}$ [3][4]

To describe the quadratrix as simple function rather than planar curve, it is advantageous to switch the y-axis and the x-axis, that is to place the side AB on y-axis rather than on the x-axis. Then the quadratrix is given by the following function f(x):[5][6]

${\displaystyle f(x)=x\cdot \cot \left({\frac {\pi }{2a}}\cdot x\right)}$

## Angle trisection

The trisection of an arbitrary angle using only ruler and compasses is impossible. However, if the quadratrix is allowed as an additional tool, it is possible to divide an arbitrary angle into n equal segments and hence a trisection (n = 3) becomes possible. In practical terms the quadratrix can be drawn with the help of a template or a quadratrix compass (see drawing).[1][2]

Since by the definition of the quadratrix the traversed angle is proportional to the traversed segment of the associated squares' side dividing that segment on the side into n equal parts yields a partition of the associated angle as well. Dividing the line segment into n equal parts with ruler and compass is possible due to the intercept theorem.

For a given angle BAE (  90°) construct a square ABCD over its leg AB. The other leg of the angle intersects the quadratrix of the square in a point G and the parallel line to the leg AB through G intersects the side AD of the square in F. Now the segment AF corresponds to the angle BAE and due to the definition of the quadratrix any division of the segment AF in n equidistant parts yields a corresponding division of the angle BAE into n parts of equal size. To divide the segment AF into n equidistant parts proceed as follows. Draw a ray a with origin in A and then draw n equidistant segments (of arbitrary length) on it. Connect the endpoint O of the last segment with F and draw lines parallel to OF through all the endpoints of the remaining n  1 segments on AO, these parallel lines divide the segment AF on AD into n equidistant segments. Now draw parallel lines to AB through the endpoints of those segments on AF, these parallel lines will intersects the trisectrix. Connecting those points of intersection with A yields a partition of angle BAE into n parts of equal size.[5]

Since not all points of the trisectrix can be constructed with circle and compass alone, it is really required as an additional tool next to compass and circle. However it is possible to construct a dense subset of the trisectrix by circle and compass, so while you cannot assure an exact division of an angle into n parts without a given trisectrix, you can construct an arbitrarily close approximation by circle and compass alone.[2][3]

## Squaring of the circle

Squaring the circle with ruler and compass alone is impossible. However, if one allows the quadratrix of Hippias as an additional construction tool, the squaring of the circle becomes possible due to Dinostratus' theorem. It lets one turn a quarter circle into square of the same area, hence a square with twice the side length has the same area as the full circle.

According to Dinostratus' theorem the quadratrix divides one of the sides of the associated square in a ratio of ${\displaystyle {\tfrac {2}{\pi }}}$.[1] For a given quarter circle with radius r one constructs the associated square ABCD with side length r. The quadratrix intersect the side AB in J with ${\displaystyle \left|{\overline {AJ}}\right|={\tfrac {2}{\pi }}r}$. Now one constructs a line segment JK of length r being perpendicular to AB. Then the line through A and K intersects the extension of the side BC in L and from the intercept theorem follows ${\displaystyle \left|{\overline {BL}}\right|={\tfrac {\pi }{2}}r}$. Extending AB to the right by a new line segment ${\displaystyle \left|{\overline {BO}}\right|={\tfrac {r}{2}}}$ yields the rectangle BLNO with sides BL and BO the area of which matches the area of the quarter circle. This rectangle can be transformed into a square of the same area with the help of Euclid's geometric mean theorem. One extends the side ON by a line segment ${\displaystyle \left|{\overline {OQ}}\right|=\left|{\overline {BO}}\right|={\tfrac {r}{2}}}$ and draws a half circle to right of NQ, which has NQ as its diameter. The extension of BO meets the half circle in R and due to Thales' theorem the line segment OR is the altitude of the right angle triangle QNR. Hence the geometric mean theorem can be applied, which means that OR forms the side of a square OUSR with the same area as the rectangle BLNO and hence as the quarter circle.[7]

Note that the point J, where the quadratrix meets the side AB of the associated square, is one of the points of the quadratrix that cannot be constructed with ruler and compass alone and not even with the help of the quadratrix compass based on the original geometric definition (see drawing). This is due to the fact, that the 2 uniformly moving lines coincide and hence there exists no unique intersection point. However relying on the generalized definition of the quadratrix as a function or planar curve allows for J being a point on the quadratrix.[8][9]

## Historical sources

The quadratrix is mentioned in the works of Proclus (412–485), Pappus of Alexandria (3rd and 4th centuries) and Iamblichus (c. 240  c. 325). Proclus names Hippias as the inventor of a curve called quadratrix and describes somewhere else how Hippias has applied the curve on the trisection problem. Pappus only mentions how a curve named quadratrix was used by Dinostratus, Nicomedes and others to square the circle. He neither mentions Hippias nor attributes the invention of the quadratrix to a particular person. Iamblichus just writes in a single line, that a curve called a quadratrix was used by Nicomedes to square the circle.[10][11][12]

Though based on Proclus' name for the curve it is conceivable that Hippias himself used it for squaring the circle or some other curvilinear figure, most historians of mathematics assume that Hippias invented the curve, but used it only for the trisection of angles. Its use for squaring the circle only occurred decades later and was due to mathematicians like Dinostratus and Nicomedes. This interpretation of the historical sources goes back to the German mathematician and historian Moritz Cantor.[11][12]

## Notes

1. Horst Hischer: Klassische Probleme der Antike – Beispiele zur "Historischen Verankerung" Archived 2012-03-28 at the Wayback Machine. In: Blankenagel, Jürgen & Spiegel, Wolfgang (Hrsg.): Mathematikdidaktik aus Begeisterung für die Mathematik — Festschrift für Harald Scheid. Stuttgart/Düsseldorf/Leipzig: Klett 2000, pp. 97 – 118
2. Hans-Wolfgang Henn: Elementare Geometrie und Algebra. Verlag Vieweg+Teubner 2003, pp. 45–48 "Die Quadratur des Kreises" (excerpt, p. 47, at Google Books)
3. Hans Niels Jahnke: A History of Analysis. American Mathematical Society 2003, ISBN 0821826239, pp. 30–31 (excerpt, p. 30, at Google Books)
4. Weisstein, Eric W. "Quadratrix of Hippias". MathWorld.
5. Dudley Underwood: The Trisectors. Cambridge University Press 1994, ISBN 0883855143, pp. 6–8 (excerpt, p. 6, at Google Books)
6. Audun Holme: Geometry: Our Cultural Heritage. Springer 2010, ISBN 9783642144400, pp. 114–116 (excerpt, p. 114, at Google Books)
7. Jean-Paul Delahaye: Pi – Die Story. Springer 1999, ISBN 3764360569, p. 71 (excerpt, p. 71, at Google Books)
8. Van der Waerden: Science Awakening. Oxford University Press 1961, p. 146
9. James Gow: A Short History of Greek Mathematics. Cambridge University Press 2010, ISBN 9781108009034, pp. 162–164 (excerpt, p. 162, at Google Books)
10. Thomas Little Heath: A History of Greek Mathematics. Volume 1. From Thales to Euclid. Clarendon Press 1921 (Nachdruck Elibron Classics 2006), pp. 182, 225–230 (online copy at archive.org)


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# Prove that the function is continuous at $x=1$

Given a function $f: \Bbb R \rightarrow \Bbb R$

$f(x) = \begin{cases} 1/x, & \text{if$x\geq1$} \\ 2-x, & \text{if$x<1$} \end{cases}$

I need to prove that this function is continuous at $x=1$ using $\varepsilon$ and $\delta$ notations

So far I did the following and then got stuck:

By definition $f(x)$ is continuous at $c$ if $\displaystyle \lim_{x\to c} f(x) = f(c)$. Hence, I will consider the following limits:

1) Claim: $\displaystyle \lim_{x\to1^-} f(x) = 1$.

For this claim to be true the following must hold: for $\varepsilon>0$ there exists $\delta>0$ such that $1-\delta<x<1$ (1) implies $|f(x) - f(1)|<\varepsilon$. Since $x$ approaches from below we take $f(x)=2-x$ and consider:

$|f(x) - f(1)| = |1-x|$

Now we set $\delta=\varepsilon$ and by subbing into (1) we get:

$1-\varepsilon<x<1$

$-\varepsilon<x-1<0<\varepsilon$

$|x-1|<\varepsilon$ which gives that $\displaystyle \lim_{x\to1^-} f(x) = 1$.

2) Claim $\displaystyle \lim_{x\to1^+} f(x) = 1$. For this claim to be true the following must hold: for $\varepsilon>0$ there exists $\delta>0$ such that $1<x<1+\delta$ (1) implies $|f(x) - f(1)|<\epsilon$. Since $x$ approaches from below we take $f(x)=\dfrac 1x$ and consider:

$|f(x) - f(1)| = \left|\dfrac 1x -1\right|= \left|\dfrac{x-1}{x}\right|$

From $1<x<1+\delta$ we can deduce that $|x-1|<\delta$ But now I am stuck and I am not sure what my $\delta$ should be for $|(x-1)/x)|<\varepsilon$

If I could prove that my second claim is true then 1) and 2) would imply $\displaystyle\lim_{x\to 1} f(x) = 1$ and so the function would be continuous.

Thank you and I hope my formatting is not that bad.

## 3 Answers

Solve the equation : \begin{align} &-\epsilon<\frac1x-1<\epsilon \\ \iff& 1-\epsilon<\frac1x<1+\epsilon \\ \iff& \frac{1}{1-\epsilon}>x>\frac{1}{1+\epsilon} \\ \iff& \frac{\epsilon}{1-\epsilon}>x-1>-\frac{\epsilon}{1+\epsilon} \end{align} As $x-1>0$, it suffices to choose $\delta=\frac{\epsilon}{1-\epsilon}$ (for $\epsilon<1$) to be able to conclude.

• Thank you! This now makes sense! – LiNXO Nov 27 '16 at 13:17

I have to say that your understanding about "x tends to c" is wrong, since you just considered one side situation. In fact, the approximating sequence can switch to different sides of 1, for example, $x_n=1\pm\frac{1}{n}$. So in fact the correct formulation should be $|x_n-1|\rightarrow 0$.

Back to your question, it should follow that $$|f(x_n)-f(1)|=|f(x_n)-1|\leq \max(|\frac{1}{x_n}-1|,|1-x_n|).$$ Let $|x_n-1|<\epsilon$. Inparticular, $x_n\neq 0$ for sufficiently small $\epsilon$. Then the first term can be estimated by $$|\frac{1}{x_n}-1|=\frac{|x_n-1|}{|x_n|}\leq\frac{\epsilon}{1+\epsilon}\leq\frac{\epsilon}{2}\leq\epsilon$$ for $\epsilon<1$; for the second term, just use the definition of $x_n$. Now you conclude that $|f(x_n)-f(1)|\leq \epsilon$ for sufficiently large $n$, and you conclude the continuity.

1) as $2-x$ is a decreasing function, you need to establish on the left

$$-\delta<x-1<0\implies f(1)-f(x)=x-1>-\epsilon,$$ which is obtained with $\delta=\epsilon$.

2) as $1/x$ is a decreasing function, you need to establish on the right

$$0<x-1<\delta\implies f(1)-f(x)=1-\frac1x<\epsilon,$$ which is obtained with $1/x>1-\epsilon$, or $x-1<\epsilon/(1-\epsilon)=\delta$.

Then

$$f(1^-)=f(1^+)=f(1).$$


text:
# Find all $n\in\mathbb{Z}^+$ such that the sum of the digits of $5^n$ equals $2^n$ [duplicate]

Find all $n\in\mathbb{Z}^+$ such that the sum of the digits of $5^n$ equals $2^n$

Starting with a table of values, I found that $n=3$ works. Beyond this, it's hard to imagine any other number working. As $n$ gets larger, it seems to me that $2^n$ grows far more rapidly than the sum of the digits of $5^n$ would ever be able to "catch up" to.

My reasoning: multiplying by each $2$ doubles $2^n$, but multiplying by each $5$ adds (I think) at most $1$ digit to $5^n$, which would add at most $9$ to the sum of the digits of $5^n$. Another way of looking at it, suppose $n=1000000$. Notice that $5^{1000000}<10^{1000000}$, which is $1$ with $1000000$ zeros. Thus $5^{1000000}$ has at most $1000000$ digits, and the sum of its digits would be maximized if those digits were all $9$s. Then the sum of the digits of $5^n$ would be $9\cdot1000000=9000000$, which is far less than the number $2^{1000000}$.

But is there a way to prove this using theory?

## marked as duplicate by JMoravitz, Ross Millikan, Jonas Meyer, Peter Woolfitt, ElaqqadMar 29 '15 at 18:07

• Sounds like you already have the theory to me. What specific difficulty are you encountering trying to write this up as a general argument? (i.e. not just about $n=1000000$). – Erick Wong Mar 29 '15 at 16:09
• Indeed, try to find the smallest $n$ such that $9\cdot n\log_{10}5 < 2^n$. In otherwords, the smallest $n$ for which the sum of digits of $5^n$ is always smaller than $2^n$ regardless what the digits actually are. – JMoravitz Mar 29 '15 at 16:10
• @ErickWong Thank you, but my thinking (to me) is just "informal" and I don't think it's acceptable as a proof. – yroc Mar 29 '15 at 16:12
• One method would be induction. Show that, if $9n<2^n$, then $9(n+1)<2^{n+1}$ as well. You also need a starting-point, as JMoravitz says. – Empy2 Mar 29 '15 at 16:13
• That is a different question @Elaqqad, where it compares sum of digits of $5^n$ to sum of digits of $2^n$, whereas in this question we are comparing sum of digits of $5^n$ to the number $2^n$ itself. – JMoravitz Mar 29 '15 at 16:23

The number of digits of $5^n$ is $\leq n$. So, their sum is $\leq 9n$.

If $n\geq6$, $2^n>9n$. Therefore, we only need to test it for $n=1,2,3,4,5$

Out of those, it is only true for $n=3$.

• @I like your answer but, in a proof, is it not necessary to justify "the number of digits of $5^n\le n$" – yroc Mar 29 '15 at 16:31
• I skipped it since I thought it was obvious. But it is possible to prove it.  The number of digits of a positive integer is an increasing function (Let's write is a $d(n)$). So $$\forall n\geq 1:d(5^n)\leq d(10^n-1)=d(10^n)-1=n+1-1=n$$ – Kitegi Mar 29 '15 at 16:47
• @yroc Note that Farnight's proof exactly follows your ad hoc argument that $5^{1000000} < 10^{1000000}$ is $1$ with $1000000$ zeroes, so it has at most $1000000$ digits. Very often the essence of writing a proof is to take something "obvious for one number" like this and distilling it so it becomes clear for all values of $n$ (or making it clear exactly which $n$ the same lline of reasoning applies to). It's a skill that I'd encourage you to cultivate. – Erick Wong Mar 29 '15 at 17:10
• @ErickWong That's good advice; thank you kindly. – yroc Mar 29 '15 at 19:56

Hint: Digital sum of a number $x$ in base $10$ is given by,

$$\sum_{n=0}^{\lfloor{\log_{10}(x)}\rfloor} \frac{1}{10^n} (x \bmod 10^{n+1} - x \bmod 10^n)$$

• @Hmm, good formula to know! Thanks. – yroc Mar 29 '15 at 16:36


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text:
# Life on a Venusian world?

I define "Venusian world" to be high pressure, thick atmosphere, and high temperatures. Assume the atmosphere is similar to Venus, but whatever gets the same effect and supports life is fine.

I'm envisioning a six-limbed sentient species that's nearly blind and uses sonar to "see", and a primarily fungal-like plant life, with possibly some air-based floating life on the top surface, though small and non-sentient.

Would it be possible to have (water) oceans in this sort of planet and would it be possible to have water/carbon-based life forms?

• I like the new question, too. Can we assume a similar atmospheric composition to Venus, or no? – HDE 226868 Jan 27 '15 at 15:34
• Hmm...similar, but it doesn't have to be exact. If life doesn't work with similar, then dissimilar is fine too. – user6383 Jan 27 '15 at 15:42
• Boiling points of water change with pressure changes...at 92 ATM (94232.25 Millibar) water boils around 495K (about 223 degrees celcius). Venus runs at about 465 degrees Celcius...if you were to reduce the heat on Venus by 265 degrees or so, liquid surface water is actually a potential. I'd imagine a much younger version of Venus may have had closer to these conditions (perhaps an older/cooler one, but Venus possesses such a strong Greenhouse effect that an older Venus may be warmer) – Twelfth Jan 27 '15 at 18:55
• How about similar in pressure, but much cooler? Say the planet is located about Mars' distance from a sunlike star? Because I don't think you are going to get anything remotely resembling Earth-like carbon chemistry working at Venusian temperatures. Though somewhat higher temperature & pressure works - see the extremophiles living around deep ocean vents - as long as you can get liquid water. – jamesqf Jan 27 '15 at 18:56
• Seems like @jamesqf had the same idea I had. A planetary scale event that relocated Venus to a much further orbit would work for this. Gravity only, no impact or anything...brown dwarf got too close and relocated it maybe? – Twelfth Jan 27 '15 at 19:09

• Surface gravity: $8.87 \text{ m/s}^2$
• Surface temperature: $737 \text{ K}$
• Surface pressure: $\approx$ $92 \text{ atm}$

So no liquid water on the surface.

Wikipedia also mentions later on that you have to get 50,000 meters in the air before conditions become something like Earth (i.e. temperature and pressure are somewhat survivable). That gives us a nice little loophole. A little searching on Wikipedia showed that someone else had had the same idea I had: floating colonies:

Now we're getting somewhere.

Once more, I go back to my pufferpolyp, a floating creature that has an air sac carrying lifting gases. It floats along on its merry way, essentially staying at the same altitude for all its life (although when it dies, there's a chance its air sac will deflate and it will fall 50,000 meters to the surface, where it will be crushed and roasted).

Anyway, my life of choice is this pufferpolyp. Unfortunately, the amount of oxygen in the Venusian atmosphere is next to nil. Instead, there's plenty of $\text{CO}_2$. That gives us a little loophole. 50,000 feet should bring you above the clouds of sulfur dioxide, meaning that you might get some sunlight. So this pufferpolyp can use photosynthesis!

After a long time, if enough pufferpolyps are around, the upper atmosphere will turn to a less hellish mixture and could become an atmosphere of oxygen and nitrogen. Will the pufferpolyps die? Maybe; maybe not. $\text{CO}_2$-breathing life on Earth survived the Great Oxygenation Event 2.3 billion years ago. I think they'll survive. Bacteria might be able to take in oxygen and put out $\text{CO}_2$.

And if the pufferpolyps have been "deployed" by another race intent on aerial colonization . . . there's going to be a lot of changes happening soon.

There's no way that something as weird as a pufferpolyp could develop in such a hellish place. Most living things aren't born floating 50,000 meters in the air, and this is no exception. The things couldn't have floated up from the surface because they'd be squished in about 10 minutes down there. So my cop-out is that they've been put there by a race (humans?) intent on terraforming the planet, starting with the upper atmosphere.

That's not to say that it's impossible for things to live and evolve in the sky. Bacteria have been found living in our atmosphere; it's somewhat plausible that a similar thing happened on this planet, but the bacteria began evolving into multicellular organisms, eventually leading to the pufferpolyps.

• @cntrational Under higher pressure, yes. – HDE 226868 Jan 27 '15 at 16:02
• Sorry, accidentally deleted my comment. It was about higher pressures (98 atm) allowing water to form. – user6383 Jan 27 '15 at 16:04
• @cntrational Yes; see, for example, this: www1.lsbu.ac.uk/water/water_phase_diagram.html#intr. – HDE 226868 Jan 27 '15 at 16:04
• The issue with something like a pufferpolyp is that you run into an evolutionary problem. You'd basically have to figure out a way for life to evolve almost entirely in the atmosphere, which, given the resource constraints and lack of solvents, is highly unlikely. – Nick2253 Jan 27 '15 at 16:08
• Alternatively, maybe the planet wasn't always Venusian. Perhaps it was originally more earthlike, life developed (including an ancestor of your pokemon), and then later it went into runaway greenhouse effects which caused the current atmosphere. But a few flying creatures and bacteria managed to survive and continue to evolve. Could have happened hundreds of millions of years ago to give the new atmosphere time to develop. – Dan Smolinske Jan 27 '15 at 18:15


text:
MathSciNet bibliographic data MR922879 11J82 (41A21) Rukhadze, E. A. A lower bound for the approximation of ${\rm ln}\,2$${\rm ln}\,2$ by rational numbers. (Russian) Vestnik Moskov. Univ. Ser. I Mat. Mekh. 1987, no. 6, 25–29, 97. Links to the journal or article are not yet available

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
## § HoTTesT: Identity types

• We already have judgemental equality. (computational equality: beta eta alpha equality).
• We will next introduce propositional equality.
• Proving an equality => constructing a term of the type of equalities.
• We can prove many judgemental equalities (eg. add x 0 =judgement= x), but not all the ones we want (eg. add 0 x =judgement= x).
• We can't because we need to do induction on x.
• When we use natural number elimination / induction, we must produce a term of a type.
• What type?!
• Type constructors internalize structure. Eg. at a meta level, we can talk about contexts x:A, y:B(x), z:C(x,y).
• But internally, we will need to be able to talk about contents, we can use sigma types! Σ(x:A) Σ(y: Bx) Σ z:C(x, y)lets us internalize contexts!
• Similarly, we can think about dependent terms as meta functions. Example, x:A,y:B(x) |- c(x,y): C(x, y). We can think of this as a function that takes an x and a y and produce a c(x,y). See that pi types internalize this notion! c: (x:A) -> (y: B(x)) -> C(x,y).
• Bool, Nat, etc. are internalizing the usual booleans, naturals, etc.
• The universe type internalizes the judement A is a type via A: Type.
• The identity type internalizes the meta notion of judgemental equality (how? Doesn't it prove strictly more?)

#### § Identity Type

• $=$-formation: A type a: A, b: B, then we have a type a =A b type.
• =-intro: a:A| r_a: a =A a. ( r = reflexivity).
• =-elim: x: A, y: A, z: x =A y |- D(x, y, z) type, and given x:A |- d: D(x, x, r_x), then we have ind=(d, x, y, z): D(x, y, z)


text:
A stone tied to the end of a string 80 cm long is whirled in a

Question.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

solution:

Length of the string, $I=80 \mathrm{~cm}=0.8 \mathrm{~m}$

Number of revolutions $=14$

Frequency, $v=\frac{\text { Number of revolutions }}{\text { Time taken }}=\frac{14}{25} \mathrm{~Hz}$

Angular frequency, $\omega=2 \pi \mathrm{V}$

$=2 \times \frac{22}{7} \times \frac{14}{25}=\frac{88}{25} \mathrm{rad} \mathrm{s}^{-1}$

Centripetal acceleration, $a_{\mathrm{c}}=\omega^{2} r$

$=\left(\frac{88}{25}\right)^{2} \times 0.8$

$=9.91 \mathrm{~m} / \mathrm{s}^{2}$

The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.


text:
Tek's Domain

#<NTA:NnT:SSrgS:H6.6-198:W200-90.72:CBWg>

TCP, UDP, and SCTP

Now I know that if you’ve been on the internet long enough, you’ve at least heard the term TCP thrown around. Maybe UDP too, and to any networking people here, you should be very familiar with what they are and how they compare. Well today, I bring the unheard third layer 4 protocol into the spotlight: SCTP

Layer 4 Protocols

In the OSI model, layer 4, or the transport layer, is what provides host-to-host communication for applications. Layer 4 should provide services such as connection-oriented communication, reliability, flow control, and multiplexing.

Obviously the most well-known L4 protocol and the one that almost everything on the internet uses is TCP (Transmission Control Protocol), though a few do use the second-place winner, UDP (User Datagram Protocol). Now just because those are the two major ones does not mean that there aren’t more, DCCP (Datagram Congestion Control Protocol), and SCTP (Stream Control Transmission Protocol)

I’ll be covering all three, because why not.

Transmission Control Protocol (TCP)

TCP, which has been around since the invention of the Internet Protocol (IP), provides reliable, ordered, and error-checked delivery of a stream of bytes, between hosts. TCP is connection-oriented, meaning you have to deliberately decide to and agree to talk to another computer, and also provides for error correction, and retransmission of lost data.

TCP packets are given a number relative to each connection, called the sequence number. This allows both computers to know which ones have been received, and which ones failed somewhere and need to be sent again.

My favorite analogy for TCP is like two gentlemen starting a discussion:

Would you like to talk about <topic>?

I would indeed like to, if you are willing.

I am willing, let us talk.

[talk ensues]

That is all I have to say on the matter.

Very well, that's all I have to say too.

Good, then this discussion is finished.

Or in the case of a host rejecting a connection:

Would you like to talk about <topic>?

No sir, indeed I would not like to talk about <topic>.

TCP is meant to withstand the fundamental assumption of the internet: that network links are unstable, and can randomly fail or lose data at any time. It can be a little slow (relatively), but almost guarantees that if you send something, it will be received.

There is one big point to note: TCP works on a byte stream: there is no concept of the “beginning” or an “end” of a message. It’s just all bytes. This is why, for example, HTTP, specifies that a double CR LF pair is what signals the end of transmission, because TCP itself does not provide a method of saying “they’re finished, your turn.”

Ports

TCP, like all three protocols here, uses the concept of ports to demultiplex connections. A “port” is just a number, in all three cases, from 1 to 65535, inclusive. Ports do not signify anything, they are just a number that can be used by the host’s networking equipment to determine what application needs to be sent the incoming data. Some ports are commonly used for services, such as 80 for HTTP, 443 is HTTPS, 143 is IMAP, 22 is SSH, and so on. All you’re telling the receiving computer is “send this to port 80.” The receiving computer sees this, and the processing can be explained as “this is a packet for port 80, and httpd has requested to use port 80, therefore I’ll deliver it to httpd.”

Flags

All TCP packets have a set of “flags”: boolean yes-or-no options used to convey some data. These are used for connection set-up, tear-down, and various other state information that needs to be transmitted to provide reliable transport (oversimplified explanation).

User Datagram Protocol (UDP)

UDP, unlike TCP, is connectionless: there is no formal agreement on communicating, you just.. fire off some data. Depending on the program doing this, the appropriate analogy is anywhere from getting hit in the face with a letter folded into a paper airplane, to just turning a fire hose of data to someone, and blasting them with it.

UDP has very minimal overhead, and doesn’t really meddle too much with your data. You have the ports and a data checksum (that’s actually optional). The downside is that, in trading off the overhead, UDP is now considered “unreliable” in the sense that if something happens to the data, there is no mechanism in place by UDP itself to correct this. There’s an (optional) checksum but that has to be verified by the program receiving it, not the networking service as part of the OS, like on TCP, where it would be corrected before being given to the app.

There’s also a few other differences:

• UDP is very simple, and can be handled by devices without a full network stack.
• It’s stateless, there is no information carried over from one message to another.
• UDP sends datagrams, or distinct groups of bytes. Unlike TCP where all messages are just concatenated together as one continuous stream of data, in UDP if they’re sent separately, they’re delivered separately, meaning that there is an actual definition for the start and end of messages.

UDP has a few uses: the Network File System protocol (NFS), as well as three core protocols of modern networks, Network Time Protocol (NTP), Dynamic Host Configuration Protocol (DHCP), and the Domain Name Service (DNS), all use UDP as their transport, not TCP.

Additionally, voice and video stream data, such as a livestream or VoIP session (Discord, anyone?), usually prefer UDP because the lack of latency, and for both those types, if one packet is not correct, it’s likely not going to be that big of a deal.

Stream Control Transmission Protocol (SCTP)

SCTP is.. well, kinda a mix of both TCP and UDP. Heck, the few libraries that I’ve seen that support it literally offer options to treat an SCTP connection like either one, with the standard calling conventions.

SCTP, like UDP, is a datagram transport, not a stream transport. Like TCP though, it provides reliable, ordered, congestion controlled transport.

SCTP does have a formal connection handshake, and is capable of breaking large messages into chunks, which can be transmitted separately and then re-assembled on the other end.

SCTP has a “stream identifier”, in each chunk, meaning that you can transmit several independent pieces of data at the same time over the same HTTP connection. This is very similar in implementation to ports, or to how HTTP/2 multiplexes data (also called streams there too).

SCTP also has a feature called “multi-homing”, where it will build multiple routes to the destination, such that if one fails, it can continue on another, thus improving reliability.

So, for all these features, what’s the downside?

Well, it does have some latency, like TCP. The main downside though, is just that very few things really support it, and almost no applications actually use it for any meaningful purpose, though some cheap “smart” devices use SCTP to force you to use their proprietary software to decode their “proprietary” protocol (no it’s not).

It’s really cool in theory, though likely not information you’re going to use in the wild. Good things to know, though, I guess?


text:
# Griffiths 4.12

1. Homework Statement
This question refers to Griffiths E and M book.

Of course I am stuck with the integral.

I chose z to be in the same direction as the polarization.

The integral is then
$$P \int d\Omega r'^2 dr'\frac{\left(\mathbf{r} -\mathbf{r'}\right)\cdot \hat{\mathbf{z}}}{\left(\mathbf{r} -\mathbf{r'}\right)^2}$$

BTW, how do you get the script r that Griffiths uses in latex?

2. Homework Equations

3. The Attempt at a Solution

Your denominator should be $$({\bf r-r')}^3$$.


text:
# How can I draw a Bézier Curve through a set number of points?

For high school Mathematics Pre-Specialist, I have been given the task of writing a mathematical investigation based on the following three questions:

Quadratic Bezier curve enables a smooth curve to be drawn through two points . In the example that you have considered the smooth curve passed through points A and C with point B being the control point that determines the shape of the curve. The parametric equations for these bezier curves will be quadratics (ie they involve the parameter squared) 1. Draw at least 3 Bézier curves for different positions of the control point B using the same end points A (-3,4) and C (4,-6) and include the curves in your report. • Show how the parametric equations for these curves were developed. • What properties do all of these curves have in common? • Explain how the position of the control point changes the shape of the curve. A Quadratic Beziér curve can be used to construct a smooth curve to connect two line segments, in this case the control point will be the intersection point of the two line segments extended. 2. Sketch the line segments given below • AB y=2x+1 0

Since we have been studying Vectors and this Bézier Curve assignment is merely a side project, I have no idea how to complete it. The questions appear to make sense, so my biggest problem is likely to be drawing a Bézier Curve through, say, two points. I've been trying to do this in Geogebra, but to no avail. How can it be done?

The great thing about Bezier curves is their simplicity. To draw a curve through $A,B,C$, you take $$g(t) = (1-t)^2 A + 2t(1-t) B + t^2 C$$ For $t = 0$, this gives $A$; for $t = 1$, you get $C$, and for $t = 0.5$, you get $A/4 + B/2 + C/4$, i.e., the average of $B$ with the midpoint of $AC$.

To make that concrete, you have to know what I mean by something like $0.2 A$. What I mean is this: If $A = (x, y)$, then by $0.2A$ I mean $(0.2 x, 0.2 y)$, i.e., term by term multiplication by the constant. You also need to know what I mean by adding points: I mean term-by-term addition.

There's a lovely geometric construction, too:

For any value $t$, you can find a point $t$ of the way from $A$ to $B$ by computing $(1-t)A + tB$. When $t = 0$, this gives $A$; when $t = 1$, it gives $B$. This is the first-order Bezier curve with two control points. The cool thing is that to compute the 2nd-order curve with three points, $A,B,C$, you do the following:

1. Go $t$ of the way from $A$ to $B$; call that $P$
2. Go $t$ of the way from $B$ to $C$; call that $Q$
3. Go $t$ of the way from $P$ to $Q$; call that $R$.

Then $R$ is what I've called $g(t)$ above. It depends on $t$; as you vary $t$ from $0$ to $1$, $R$ traces out a Bezier curve from $A$ to $C$, passing near $B$.

Here's an alternative that's quite surprising (to me at least): Starting with a two-segment polyline, from $A$ to $B$ to $C$, you can do the first two parts of the construction above for $t = 0.5$, so that $P$ is the midpoint of $AB$, and so on. You end up with six points: $A, B, C, P, Q, R$. And from these, you can draw a polyline through $A, P, R, Q, C$, i.e., a four-segment polyline.

Now considering $A, P, R$ and $R, Q, C$ as starting points, and do this again (i.e., treat each group as the control points for a Bezier curve, and apply the $t = 0.5$ subdivision to it to get a polyline with even shorter edges. You end up with 8 edges. Repeat again, and again...and the results will converge to a set of points on the "true" quadratic Bezier curve defined by $t \mapsto g(t)$.


text:
# Frequency-based Multi Task learning With Attention Mechanism for Fault Detection In Power Systems

15 Sep 2020  ·  , ·

The prompt and accurate detection of faults and abnormalities in electric transmission lines is a critical challenge in smart grid systems. Existing methods mostly rely on model-based approaches, which may not capture all the aspects of these complex temporal series. Recently, the availability of data sets collected using advanced metering devices, such as Micro-Phasor Measurement units ($\mu$ PMU), which provide measurements at microsecond timescale, boosted the development of data-driven methodologies. In this paper, we introduce a novel deep learning-based approach for fault detection and test it on a real data set, namely, the Kaggle platform for a partial discharge detection task. Our solution adopts a Long-Short Term Memory architecture with attention mechanism to extract time series features, and uses a 1D-Convolutional Neural Network structure to exploit frequency information of the signal for prediction. Additionally, we propose an unsupervised method to cluster signals based on their frequency components, and apply multi task learning on different clusters. The method we propose outperforms the winner solutions in the Kaggle competition and other state of the art methods in many performance metrics, and improves the interpretability of analysis.

PDF Abstract

## Code Add Remove Mark official

No code implementations yet. Submit your code now


text:
A spherical naphthalene ball of $2\:mm$ diameter is subliming very slowly in stagnant air at $25^{\circ}C$. The change in the size of the ball during the sublimation can be neglected. The diffusivity of naphthalene in air at $25^{\circ}C$ is $1.1\times10^{-6}\:m^2/s$.

The value of mass transfer coefficient is $B\times10^{-3}\:m/s$, where $B$ (up to one decimal place) is ____________ .


text:
## Calculus 8th Edition

Maximum: $f(-2,\sqrt{12}),f(-2,-\sqrt{12})=47$ and Minimum: $f(1,0)=-7$
Our aim is to calculate the extreme values with the help of Lagrange Multipliers Method subject to the given constraints. For this, we have:$\nabla f(x,y)=\lambda \nabla g(x,y)$ This yields $\nabla f(x,y)=\lt 4x-4,6y \gt$ and $\lambda g(x,y)= \lt x,y \gt$ Using the constraint condition $x^2+y^2 \leq 16$ we get, $x^2=\pm \dfrac{9}{2}$ and $x= \pm \dfrac{3}{\sqrt 2}$ $\nabla f(x,y)=\lt 4x-4,6y \gt$ and $\lambda g(x,y)= \lt x,y \gt$ we get $x=-2$ and $y=\pm \sqrt{12}$ Hence, Maximum: $f(-2,\sqrt{12}),f(-2,-\sqrt{12})=47$ and Minimum: $f(1,0)=-7$


text:
# Minimal Universal Exponent

Last updated 2019-05-19

This was a project for a number theory course: MATH 4150 at Georgia Tech. The goal was to compute the minimal universal exponent for a number, also known as the Carmichael function.

The minimal universal exponent of a positive integer $n$ is the smallest positive integer $m$ such that:

$a^m \equiv 1 \pmod n \quad\forall a \in \Z_n^+ \quad\textrm{where}\quad \textrm{gcd}(a, n) = 1$

This Carmichael function, we’ll refer to it as $\lambda(n) = m$, is instrumental to the generation of RSA keys.

In order to compute the function, we have to break down $n$ to it’s prime factors. Then by the Chinese Remainder Theorem, we can compute the minimal universal exponent using the least common multiple of the prime factors’ exponents.

$\lambda(n) = \textrm{lcm}(\lambda(p_1^{r_1}, \mathellipsis, \lambda(p_k^{r_k})) \quad\textrm{where}\quad n = \prod_{i=1}^kp_i^{r_i}$

Then for any prime factor $\lambda(p_i^{r_i})$ corresponds with Euler’s totient for that prime factor.

This means that the majority of the work by far is in factoring $n$ since computing the lcm and Euler’s totient for the prime factors is trivial. This is evident in the Chrome DevTools profile:

For reference, here is the function for computing the minimal universal exponent:

function min_unv_exp(num) {
/* Returns a number representing the minimal universal exponent
* of a positive integer num */

var e = 1; /* This will be the returned exponent, note: needs to be
one for the first lcm to work */

/* Iterate through prime factors, computing lcm of each totient */
prime_factorization(num).forEach(function(factor) {
if (factor.p === 2) {
if (factor.e == 2) {
e = lcm(e, 2);
} else if (factor.e > 2) {
e = lcm(e, Math.pow(2, factor.e - 2));
}
} else {
e = lcm(e, Math.pow(factor.p, factor.e - 1) * (factor.p - 1));
}
});

return e;
}

Digging into the prime factorization function the fastest way I could implement it was using a accumulator approach:

function prime_factorization(n) {
/* Returns an array of prime factors (in least sorted order)
* where each factor is an object of the form
* {'p': 7, 'e': 4}
* where p is the prime base and e is the exponent for p in the factorization
*/
var factors = [];
for (var p = 2; p <= n; p++) {
var e = 0;
while (n % p == 0) {
n = n / p;
e++;
}
if (e > 0) {
factors.push({ p: p, e: e });
}
}
return factors;
}

This relies on the observation that dividing $n$ by its the smallest prime factor $p_1^{r_1}$ the smallest number greater than $p_1$ that divides $n/p_1^{r_1}$ is necessarily a prime factor of $n$.

A quick proof for this follows from contradiction. Let $p_1^{r_1}$ be the factor of $n$ such that $p_1$ is minimal and $r_1$ is maximal.

Now assume $q > p_1$ is the next smallest divisor of $n$ and $q$ is not prime.

Since $q$ is composite, there must exist a prime factor $\phi < q$.

$\phi|q \,\rarr\, \phi|n$

Since $\phi|n \,\rarr\, \phi > p_1$ otherwise $p_1$ would not be the smallest prime factor of n.

However, since $q > \phi > p_1$ and $\phi|n$ that means $q$ is not the next smallest divisor of $n$ unless $q=\phi$ in which case $q$ is prime breaking our assumption that $q$ is composite. ∎


text:
# Show that $C[a,b]$ is a complete space under the metric $d(f,g)=\sup_{t\in [a,b]}|f(t)-g(t)|$.

$C[a,b]$ is a normed vector space of all continuous complex valued functions on $[a,b]$, with supremum norm

$$\|f\|_\infty=\sup_{t\in [a,b]}|f(t)|.$$

The metric induced by the norm is $$d(f,g)=\|f-g\|_\infty = \sup_{t\in [a,b]}|f(t)-g(t)|.$$

Show that $C[a,b]$ is a complete metric space under this metric induced by the supremum norm (i.e. show $C[a,b]$ is a Banach space).

• HINT: Set up a Cauchy sequence under that norm, then try to find where the limit is, see if that limit is continuous and has a finite supremum. – Shuhao Cao Oct 28 '13 at 1:59

If $\langle f_n\rangle$ is Cauchy then for any $\varepsilon >0$ there exists $N$ such that $n,m\geqslant N$ gives $$\sup_{x\in[a,b]}|f_n(x)-f_m(x)|<\varepsilon$$
But then the sequence of complex numbers $\langle f_n(x)\rangle$ is Cauchy for each $x\in [a,b]$, thus it converges. This means we can define $f:[a,b]\to \Bbb C$ by $$f(x)=\lim\limits_{n\to\infty}f_n(x)$$
It remains to show that $f_n\to f$ uniformly, and that $f$ is continuous. Can you do this?
• "$\langle f_n(x)\rangle$ is Cauchy for each $x\in [a,b]$, thus it converges." But isn't this what we have to show? – MathsMy Oct 28 '13 at 2:18
• Oh, if $f_n \rightarrow f$ uniformly then $f$ is continuous right? – MathsMy Oct 28 '13 at 2:19
• In Complex numbers, convergence $\iff$ Cauchy. But in other sets, isn't it Convergence $\rightarrow$ Cauchy but not the other way? – MathsMy Oct 28 '13 at 2:25
• @Dosomemaths Sure. But you're assuming your function goes into $\Bbb C$. – Pedro Tamaroff Oct 28 '13 at 2:26


text:
mersenneforum.org P!=NP in the news
 Register FAQ Search Today's Posts Mark Forums Read

 2010-08-10, 01:12 #12 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×11×457 Posts There are some minor word changes and a ~200-word paragraph added before Remark 7.8.
2010-08-10, 01:20   #13
joblack

Oct 2008
n00bville

10111000002 Posts

Quote:
 Originally Posted by R.D. Silverman The wording of the announcement makes it clear that this is not the effort of a professional. (too much hyperbole and informality) It is very unlikely to be correct.
LOL - always the same unmannered Silverman behavior. Heise News Service has written that he is quite established in his field so Silvermans statements have, as always, no serious impact on reality.

2010-08-10, 01:28   #14
CRGreathouse

Aug 2006

5,987 Posts

Quote:
 Originally Posted by joblack LOL - always the same unmannered Silverman behavior. Heise News Service has written that he is quite established in his field so Silvermans statements have, as always, no serious impact on reality.
At least we know what he really thinks.

But if you look at his more recent posts, Silverman has softened considerably on the paper.

2010-08-10, 01:35   #15
joblack

Oct 2008
n00bville

25·23 Posts

Quote:
 Originally Posted by CRGreathouse At least we know what he really thinks. But if you look at his more recent posts, Silverman has softened considerably on the paper.
He had no choice after it was apparent that the paper is more than 'a bad introduction'.

 2010-08-10, 01:43 #16 ATH Einyen     Dec 2003 Denmark 2·17·101 Posts The author Vinay Deolalikar writes that this version is preliminary version and was not meant for the public and final version is coming soon. http://www.hpl.hp.com/personal/Vinay_Deolalikar/ "BASIC RESEARCH Vinay Deolalikar. P is not equal to NP. 6th August, 2010 (66 pages 10pt, 102 pages 12pt). Manuscript sent on 6th August to several leading researchers in various areas. Confirmations began arriving 8th August early morning. The preliminary version made it to the web without my knowledge. I have made minor updates, here. Please note that the final version of the paper is under preparation, and is to be posted here very shortly. Stay tuned. "
2010-08-10, 04:53   #17
davieddy

"Lucan"
Dec 2006
England

11001010010102 Posts
Just fancy that...

(As Private Eye magazine would say):

Quote:
 Originally Posted by R.D. Silverman The wording of the announcement makes it clear that this is not the effort of a professional. (too much hyperbole and informality) It is very unlikely to be correct.
Quote:
 Originally Posted by R.D. Silverman I have started reading it in details. The general approach is very creative and seems to be workable.

 2010-08-10, 07:02 #18 only_human     "Gang aft agley" Sep 2002 2×1,877 Posts Some preliminary analysis and organization by Dick Lipton and Ken Regan: http://rjlipton.wordpress.com/2010/0...-p%E2%89%A0np/ . This enumerates four possible issues with the proof for ease of discussion. Last fiddled with by only_human on 2010-08-10 at 07:29 Reason: trimmed a bit
2010-08-10, 14:54   #19
only_human

"Gang aft agley"
Sep 2002

2×1,877 Posts
Wiki

http://michaelnielsen.org/polymath1/..._P!%3DNP_paper
Quote:
 Deolalikar's P!=NP paper From Polymath1Wiki Note: This is currently an UNOFFICIAL page on Deolalikar's P!=NP paper, and is not yet affiliated with a Polymath project. This is a clearinghouse wiki page for the analysis of Vinay Deolalikar's recent preprint claiming to prove that P != NP, and to aggregate various pieces of news and information about this paper. Corrections and new contributions to this page are definitely welcome. Of course, any new material should be sourced whenever possible, and remain constructive and objectively neutral; in particular, personal subjective opinions or speculations are to be avoided. This page is derived from an earlier collaborative document created by Suresh Venkatasubramanian. For the latest discussion on the technical points of the paper, see this thread of Dick Lipton and Ken Regan. For meta-discussion of this wiki (and other non-mathematical or meta-mathematical issues), see this thread of Suresh Venkatasubramanian

2010-08-10, 23:19   #20
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

22·1,877 Posts

Quote:
 Originally Posted by only_human http://michaelnielsen.org/polymath1/..._P!%3DNP_paper
Charanjit Jutla
August 10, 2010 9:50 am

Serious (unfixable) problem.

Vinay uses FO(LFP) as his main tool. However, I think
he fails to realize that FO(LFP) , although defined using
least fixed point, also contains greatest fixed points.

Thus, if \mu x f(y,x) is the formula denoting least fixpont of f(y,x) under
x,
and \nu x f(y,x) is the formula denoting greatest fixpont of f(y,x0 undex x,

then there is this triple negation rule:

\neg \mu x \neg f(y, \neg x) = \nu x f(y,x). (See Kozen's
paper on axiomatizing Mu-Calculus, or my paper with
Allen Emerson on Mu-Calculus in FOCS 1991 titled
"Mu-Calcus, Tree Automata and Determinacy", or just
original Vardi and Immerman papers). Note that the left
hand side is well defined in LFP, as x is under even
number of negations from the operator \mu.

Now, his whole theory of least fixed point only increasing
the size of the structures falls apart. BTW, this is a usual
mistake most people new to fixed-point logics fall prey to.

For example, now he has to deal with formulas of the kind
\nu x (f(y, x) \and g(y, x)).

His section 4.2 deals with just one least fixed point operator.
where his idea is correct.

But, in the next section 4.3, where he deals with complex
fixed point iterations, he is just hand waving, and possibly
way off. Given that he does not even mention greatest fixed
points leads me to think that this is a huge (unfixable) gap.

Charanjit Jutla
IBM T.J. Watson

 2010-08-11, 01:14 #21 only_human     "Gang aft agley" Sep 2002 2×1,877 Posts I was fascinated watching the developing evaluation of the paper. A preliminary consensus seems complete. Here is a blog that takes an optimistic view of this new form of peer review: http://cameronneylon.net/blog/p-%E2%...f-peer-review/ Perhaps blitz-read is a more apt term. Among my peripatetic browsing about all of this I came across Leslie Valiant's theory of holographic algorithms which exponentially speed up some computations; this was very interesting to me. There is a (mod 7) operation that seems to be part of canceling a large part of the calculations -- leaving a simpler system to be counted (and that by some kind of weighing). It reminded me a bit of the AKS method so simplifying a row.
 2010-08-12, 05:11 #22 AntonVrba     Jun 2005 2×72 Posts Now also in the BBC news: http://www.bbc.co.uk/news/science-environment-10938302

 Similar Threads Thread Thread Starter Forum Replies Last Post gd_barnes Conjectures 'R Us 305 2023-01-21 13:53 gd_barnes No Prime Left Behind 258 2023-01-21 11:09 Cruelty Riesel Prime Search 41 2010-03-08 18:46 NBtarheel_33 Hardware 17 2009-05-04 15:52 KEP Riesel Base 3 Attack 4 2008-12-17 11:54

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text:
# Environment.CollisionModels module¶

Define the different collision models.

Collision models are generic functions, taking:

• the time: t

• the arms of the current environment: arms

• the list of players: players

• the numpy array of their choices: choices

• the numpy array to store their rewards: rewards

• the numpy array to store their pulls: pulls

• the numpy array to store their collisions: collisions

As far as now, there is 4 different collision models implemented:

Environment.CollisionModels.onlyUniqUserGetsReward(t, arms, players, choices, rewards, pulls, collisions)[source]

Simple collision model where only the players alone on one arm samples it and receives the reward.

• This is the default collision model, cf. [[Multi-Player Bandits Revisited, Lilian Besson and Emilie Kaufmann, 2017]](https://hal.inria.fr/hal-01629733).

• The numpy array ‘choices’ is increased according to the number of users who collided (it is NOT binary).

Environment.CollisionModels.defaultCollisionModel(t, arms, players, choices, rewards, pulls, collisions)

Simple collision model where only the players alone on one arm samples it and receives the reward.

• This is the default collision model, cf. [[Multi-Player Bandits Revisited, Lilian Besson and Emilie Kaufmann, 2017]](https://hal.inria.fr/hal-01629733).

• The numpy array ‘choices’ is increased according to the number of users who collided (it is NOT binary).

Environment.CollisionModels.onlyUniqUserGetsRewardSparse(t, arms, players, choices, rewards, pulls, collisions)[source]

Simple collision model where only the players alone on one arm samples it and receives the reward.

• This is the default collision model, cf. [[Multi-Player Bandits Revisited, Lilian Besson and Emilie Kaufmann, 2017]](https://hal.inria.fr/hal-01629733).

• The numpy array ‘choices’ is increased according to the number of users who collided (it is NOT binary).

• Support for player non activated, by choosing a negative index.

Environment.CollisionModels.allGetRewardsAndUseCollision(t, arms, players, choices, rewards, pulls, collisions)[source]

A variant of the first simple collision model where all players sample their arm, receive their rewards, and are informed of the collisions.

Note

it is NOT the one we consider, and so our lower-bound on centralized regret is wrong (users don’t care about collisions for their internal rewards so regret does not take collisions into account!)

• This is the NOT default collision model, cf. [Liu & Zhao, 2009](https://arxiv.org/abs/0910.2065v3) collision model 1.

• The numpy array ‘choices’ is increased according to the number of users who collided (it is NOT binary).

Environment.CollisionModels.noCollision(t, arms, players, choices, rewards, pulls, collisions)[source]

Simple collision model where all players sample it and receive the reward.

• It corresponds to the single-player simulation: each player is a policy, compared without collision.

• The numpy array ‘collisions’ is not modified.

Environment.CollisionModels.rewardIsSharedUniformly(t, arms, players, choices, rewards, pulls, collisions)[source]

Less simple collision model where:

• The players alone on one arm sample it and receive the reward.

• In case of more than one player on one arm, only one player (uniform choice) can sample it and receive the reward. It is chosen by the base station.

Note

it can also model a choice from the users point of view: in a time frame (eg. 1 second), when there is a collision, each colliding user chose (uniformly) a random small time offset (eg. 20 ms), and start sensing + emitting again after that time. The first one to sense is alone, it transmits, and the next ones find the channel used when sensing. So only one player is transmitting, and from the base station point of view, it is the same as if it was chosen uniformly among the colliding users.

Environment.CollisionModels.closerUserGetsReward(t, arms, players, choices, rewards, pulls, collisions, distances='uniform')[source]

Simple collision model where:

• The players alone on one arm sample it and receive the reward.

• In case of more than one player on one arm, only the closer player can sample it and receive the reward. It can take, or create if not given, a distance of each player to the base station (numbers in [0, 1]).

• If distances is not given, it is either generated randomly (random numbers in [0, 1]) or is a linspace of nbPlayers values in (0, 1), equally spacen (default).

Note

This kind of effects is known in telecommunication as the Near-Far effect or the Capture effect [Roberts, 1975](https://dl.acm.org/citation.cfm?id=1024920)

Environment.CollisionModels.collision_models = [<function onlyUniqUserGetsReward>, <function onlyUniqUserGetsRewardSparse>, <function allGetRewardsAndUseCollision>, <function noCollision>, <function rewardIsSharedUniformly>, <function closerUserGetsReward>]

List of possible collision models

Environment.CollisionModels.full_lost_if_collision = {'allGetRewardsAndUseCollision': True, 'closerUserGetsReward': False, 'noCollision': False, 'onlyUniqUserGetsReward': True, 'onlyUniqUserGetsRewardSparse': True, 'rewardIsSharedUniformly': False}

Mapping of collision model names to True or False, to know if a collision implies a lost communication or not in this model


text:
# apple mail > inställningar - La Buona Terra

implicit interest - Swedish translation – Linguee

An implicit method has been presented for solving singular initial value problems. The method is simple and gives more accurate solution than the implicit Euler method as well as the second order implicit Runge-Kutta (RK2) (i.e., implicit midpoint rule) method for some particular singular problems. $\begingroup$ There is a difference between implicit ODE $0=F(t,y,\dot y)$ and implicit numerical methods. Most numerical methods, explicit as well as implicit, are for explicit ODE $\dot y=f(t,y)$.

Skickas inom 10-15 vardagar. Köp Moving Particle Semi-implicit Method av Seiichi Koshizuka på Bokus.com. Pris: 1009 kr. E-bok, 2018.

## Acceleration of Compressible Flow - AVHANDLINGAR.SE

Sometimes, stating something clearly can be the best way to make your  25 Nov 2020 The IT method may be regarded as an implicit version of the HOS method since it replaces the explicit perturbation approach for obtaining the still  An implicit function is one that is not defined explicitly, but the given information implies that there is a function. There is no way to solve for or explicitly. Explicit means direct or clear.

### Untitled

Implicit teaching The opposite of making facts, concepts and procedures explicit is to leave some of them implicit. This is what happens when we learn through discovery and inference. So ‘implicit teaching’ might be a good term to describe this approach, yet the expression does not seem to have caught on. The process that we used in the second solution to the previous example is called implicit differentiation and that is the subject of this section. In the previous example we were able to just solve for y y and avoid implicit differentiation. Analysis of the scheme We expect this implicit scheme to be order (2;1) accurate, i.e., O( x2 + t). Substitution of the exact solution into the di erential equation will demonstrate the consistency of the scheme for the inhomogeneous heat equation and give the accuracy.

The SQ-based collision detection method was compared to the GJK algorithm they could leverage the explicit and implicit representations of the SQ objects,  In this talk, I describe the context and method for a study that we have been conducting on the effects of blame on expressions of implicit bias.* I will provide an  Often, implicit bias can be the root of the problem. Simulation-based training is a proven method to change behaviors and improve outcomes, as well as to  Explicit or Implicit Grammar? - Grammar Teaching Approaches in Three fulltext. Jakobsson, Ina; Knutsson, Emmalinn : Malmö universitet/Lärande och  Part IV. 2004-10-20 Per Martin-Löf: Normalization by evaluation and by the method of computability.

3. The discretization of the heat equation  MPM method is also included, followed by a description of the implicit and explicit methods. A complete description of the implicit equations, both linear and  Implicit Method for the Wave Equation*. By Melvyn Ciment** and Stephen H. Leventhal***. Abstract.

The influence of a perturbation is felt immediately throughout the complete region. Crank-Nicolson Method Crank-Nicolson splits the difference between Forward and Backward difference schemes. In A popular method for discretizing the diffusion term in the heat equation is the Crank-Nicolson scheme. It is a second-order accurate implicit method that is defined for a generic equation y ′ = f(y, t) as: yn + 1 − yn Δt = 1 2(f(yn + 1, tn + 1) + f(yn, tn)).
Feministiskt tankande och sociologi

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### Getting around static typing in Scala - blog.

If there is no implicit value of the right type in scope, it will not compile. condition allows the implicit method to use arbitrarily large k, to maintain accuracy we still need k˘h2. Crank-Nicholson Method Now that we have two di erent methods for solving parabolic equation, it is natural to ask, \can we improve by taking an average of the two methods?" The answer is yes. 2021-03-28 · It is implicit in time and can be written as an implicit Runge–Kutta method, and it is numerically stable. The method was developed by John Crank and Phyllis Nicolson in the mid 20th century.


text:
# Determining R2 based on TSSEM Outputs in metaSEM

3 posts / 0 new
Offline
Joined: 06/15/2012 - 02:29
Determining R2 based on TSSEM Outputs in metaSEM

Greetings,

I have run a random-effects TSSEM (tssem2()) in metaSEM and received the estimated path loadings and correlations among factors in the output for my structural model. I also need to find the R2 values for my endogenous variables in my structural model. I was wondering how I can determine R2 values. I think they should be calculated as 1-var of the endogenous variables found in impliedS1 matrix in mx.fit component of the tssem2() output:

random2<-tssem2(...)
random2$mx.fit$impliedS1

        [,1]               [,2]            [,3]             [,4]               [,5]            [,6]             [,7]            [,8]

[1,] 1.00000000 -0.2124499 -0.37657180 -0.1408749 0.1566710 0.03624335 -0.16619094 -0.06419929
[2,] -0.21244990 0.4783484 0.57574642 0.4799511 0.5499434 0.44504316 0.53299814 0.14718394
[3,] -0.37657180 0.5757464 0.77637611 0.5277954 0.6590216 0.60979639 0.56536845 0.04923937
[4,] -0.14087491 0.4799511 0.52779536 0.5113916 0.5535190 0.40219453 0.58033105 0.22417971
[5,] 0.15667096 0.5499434 0.65902158 0.5535190 1.0000000 0.46021062 0.56435678 -0.10231421
[6,] 0.03624335 0.4450432 0.60979639 0.4021945 0.4602106 1.00000000 0.45487242 0.20794375
[7,] -0.16619094 0.5329981 0.56536845 0.5803310 0.5643568 0.45487242 1.00000000 -0.08405294
[8,] -0.06419929 0.1471839 0.04923937 0.2241797 -0.1023142 0.20794375 -0.08405294 1.00000000

Then, for example,
R2 for [,2]= 1- 0.4783484= 0.5216516.

Am I right?

I appreciate any help,

Thank you,
Hamed

Offline
Joined: 10/08/2009 - 22:37

Hi, Hamed.

In your example, 0.4783484 is already the explained variance while 0.5216516 is the error variance.

An alternative approach to calculate the error variance is diag(mxEval(Ematrix, random2$mx.fit)) . Thus, the R squares are 1-diag(mxEval(Ematrix, random2$mx.fit)) .

Cheers,
Mike

Offline
Joined: 06/15/2012 - 02:29
R2

Thank you, Mike. Your explanations helped a lot.

Cheers,
/Hamed


text:
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# IT/Mgmnt Consultant with 90 mos WorkEx requesting an Eval

Author Message
Intern
Joined: 27 Feb 2012
Posts: 14
Location: United States
Concentration: Strategy, Entrepreneurship
GMAT 1: 720 Q48 V40
GPA: 2.5
WE: Information Technology (Consulting)
Followers: 0

Kudos [?]: 5 [0], given: 2

IT/Mgmnt Consultant with 90 mos WorkEx requesting an Eval [#permalink]

### Show Tags

25 Mar 2012, 12:36
1
This post was
BOOKMARKED
Hi,

Thank you in advance for taking out the time to evaluate my profile. I am sharing below my background so you can
i) evaluate my profile,
ii) suggest target schools and backup schools (I have a few in mind)
iii) identify strong areas and improvement areas (and subsequently how to address them)

I saw the below format adopted by one of the fellow members on this forum and that it helped understand the profile better. Hence sticking to the same format.

a) Work Experience: I have 90 months of work experience in the IT and Consulting area. I currently work for one of the Big 3 IT and Consultancy companies and am based out of US. I am a Project/Program manager and lead the Portfolio management office (PMO) for a \$40 MN account. My consulting experience revolves around Project/Program management and also setting up Operations/PMO offices from scratch. In the past, I have successfully delivered complex programs running into millions of $$and have also managed the Program Office of the CSO of one of the most reputed IT consultancies. b) International Exposure: I have been based in the US for the past 2 years handling multicultural clients. I have also traveled for projects in the UK and have spent 1 year meeting and handling (by way of conf calls) clients in UK. I have spent 3-4 months in France while I was an exchange student and have traveled across Europe during my leisure time. c) GMAT: I had taken the GMAT once in 2004 and scored around 660-680 range. Since it was more than 5 years ago, I dont think it counts. I am taking the GMAT again in a month from now. d) Education: I have -a Post Graduate Diploma in Business Management and -a Bachelors in Electronics Engineering. In my PGDBM I scored around a 6.ish GPA and in my Engineering, I secured a 1st class (60%-70%) in Mumbai university, which is considered extremely strict in scoring.(I dont know how to compare/present this in the light of other universities that may not be as stringent in thier scoring) I am also a -certified Project Management Professional (PMP), -LOMA FLMI (L1) certified and -a Certified Six Sigma Black Belt (though a non practising one). e) Activities outside Profession: -I have been the Governance lead and an active organizing volunteer for the CSR wing of my organization from 2008-2010 during which I have organized many field trips to over 5 different projects that our NGO works with. I had laid out the governance model of the CSR and streamlined reporting, financials etc in those 2 years. (I hope the fact that the CSR was my organization's social work, will not make it count under my professional activities. There was no compulsion to join and it was purely a voluntary choice). -I was the Finance-in-charge and Sponsorship lead for the College committee for 2 years in a row. -I led the Jamming (music) group at my organization in the years 2009-2010 where I brought together as many as 20 people every week and we played/created music. -I have won umpteen number of singing competitions (solo and duet) and a number of Instrumental competitions. I am trained in Indian classical music and can play 5 different musical instruments (only 1 of which I have recd formal training on, rest are self-learnt). 2 of those instruments, I have played and can still play at professional levels. -I have led my college's musical club for 2 years. I was also a part of a band that was formed when I was in college where we created fusion music. -I have represented my college as its contingent leader at many inter collegiate events. f) Programs considered: I am very keen on a 1 year course like that of INSEAD, Kellogg 1Y. I am looking to start the program asap (Eg. starting Jan 2013) I am 30/M/Married and with a kid, dont know if that matters at all. Kindly do share a frank evaluation of my profile focusing on which areas need to be strengthened. Again, thank you very much for your help, support and time. Hope to hear from you soon. MBA Admissions Consultant Joined: 23 Dec 2010 Posts: 668 Followers: 12 Kudos [?]: 14 [0], given: 0 Re: IT/Mgmnt Consultant with 90 mos WorkEx requesting an Eval [#permalink] ### Show Tags 28 Mar 2012, 01:40 sjaianand wrote: Hi, Thank you in advance for taking out the time to evaluate my profile. I am sharing below my background so you can i) evaluate my profile, ii) suggest target schools and backup schools (I have a few in mind) iii) identify strong areas and improvement areas (and subsequently how to address them) I saw the below format adopted by one of the fellow members on this forum and that it helped understand the profile better. Hence sticking to the same format. a) Work Experience: I have 90 months of work experience in the IT and Consulting area. I currently work for one of the Big 3 IT and Consultancy companies and am based out of US. I am a Project/Program manager and lead the Portfolio management office (PMO) for a 40 MN account. My consulting experience revolves around Project/Program management and also setting up Operations/PMO offices from scratch. In the past, I have successfully delivered complex programs running into millions of$$ and have also managed the Program Office of the CSO of one of the most reputed IT consultancies.

b) International Exposure: I have been based in the US for the past 2 years handling multicultural clients. I have also traveled for projects in the UK and have spent 1 year meeting and handling (by way of conf calls) clients in UK. I have spent 3-4 months in France while I was an exchange student and have traveled across Europe during my leisure time.

c) GMAT: I had taken the GMAT once in 2004 and scored around 660-680 range. Since it was more than 5 years ago, I dont think it counts. I am taking the GMAT again in a month from now.

d) Education:
I have
-a Bachelors in Electronics Engineering.
In my PGDBM I scored around a 6.ish GPA and in my Engineering, I secured a 1st class (60%-70%) in Mumbai university, which is considered extremely strict in scoring.(I dont know how to compare/present this in the light of other universities that may not be as stringent in thier scoring)
I am also a
-certified Project Management Professional (PMP),
-LOMA FLMI (L1) certified and
-a Certified Six Sigma Black Belt (though a non practising one).

e) Activities outside Profession:
-I have been the Governance lead and an active organizing volunteer for the CSR wing of my organization from 2008-2010 during which I have organized many field trips to over 5 different projects that our NGO works with. I had laid out the governance model of the CSR and streamlined reporting, financials etc in those 2 years.
(I hope the fact that the CSR was my organization's social work, will not make it count under my professional activities. There was no compulsion to join and it was purely a voluntary choice).
-I was the Finance-in-charge and Sponsorship lead for the College committee for 2 years in a row.
-I led the Jamming (music) group at my organization in the years 2009-2010 where I brought together as many as 20 people every week and we played/created music.
-I have won umpteen number of singing competitions (solo and duet) and a number of Instrumental competitions.
I am trained in Indian classical music and can play 5 different musical instruments (only 1 of which I have recd formal training on, rest are self-learnt). 2 of those instruments, I have played and can still play at professional levels.
-I have led my college's musical club for 2 years. I was also a part of a band that was formed when I was in college where we created fusion music.
-I have represented my college as its contingent leader at many inter collegiate events.

f) Programs considered: I am very keen on a 1 year course like that of INSEAD, Kellogg 1Y. I am looking to start the program asap (Eg. starting Jan 2013)

I am 30/M/Married and with a kid, dont know if that matters at all. Kindly do share a frank evaluation of my profile focusing on which areas need to be strengthened. Again, thank you very much for your help, support and time. Hope to hear from you soon.

You are very experienced and successful, so most of the full-time American programs are going to assume that they don't have much to teach you (ie., you aren't a good fit for their programs). Since you are based in the US, I highly recommend you consider some of the excellent EMBA and part-time programs, depending on where you live. Yes, pursuing these studies while in a full-time job is difficult, but they might put the best-ranked/known programs in your sights. You may certainly also consider the one-year full-time Sloan Fellows/Masters programs at MIT, Stanford, or LBS or a joint program like the Global EMBA Americas and Europe.
For traditional full-time MBA programs, I do believe your chances are higher at the European programs, which are more lenient about age. If you can get your GMAT to 700 or higher, I do think you're a good candidate for INSEAD as well as IE, IESE, ESADE, and Cranfield, for example.
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Jennifer Bloom, CPRW
jbloom@accepted.com
310-815-9553

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Intern
Joined: 27 Feb 2012
Posts: 14
Location: United States
Concentration: Strategy, Entrepreneurship
GMAT 1: 720 Q48 V40
GPA: 2.5
WE: Information Technology (Consulting)
Followers: 0

Kudos [?]: 5 [0], given: 2

Re: IT/Mgmnt Consultant with 90 mos WorkEx requesting an Eval [#permalink]

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06 May 2012, 06:27
Hello,

Hope you are doing great! I just finished my GMAT yesterday and got a score of 720 (Q48, V40). Just wanted to keep you informed.

Many thanks.
Joined: 23 Dec 2010
Posts: 668
Followers: 12

Kudos [?]: 14 [0], given: 0

Re: IT/Mgmnt Consultant with 90 mos WorkEx requesting an Eval [#permalink]

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10 May 2012, 12:57
sjaianand wrote:
Hello,

Hope you are doing great! I just finished my GMAT yesterday and got a score of 720 (Q48, V40). Just wanted to keep you informed.

Many thanks.

So your options are open! Congratulations on the solid score.
_________________

Regards,
Jennifer Bloom, CPRW
jbloom@accepted.com
310-815-9553

Subscribe to Accepted's Blog

Re: IT/Mgmnt Consultant with 90 mos WorkEx requesting an Eval   [#permalink] 10 May 2012, 12:57
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