text:
Sunday, September 18, 2011

Peter Max Inspired Statue of Liberty

The above pieces were created by children 5-9

Tempera with Charcoal

Chalk Pastel and Charcoal

Step One - Pastel

Last week I saw these wonderful Peter Max inspired Statue of Liberty paintings on We Heart Art. They are full of energy and personality. I tried two versions, one with chalk pastel and the other, tempera. I laid down the chalk pastel in bright colors smearing with my finger but not being concerned about how the statue would play out on top. I took Joanna's cue with the paint version and used only primary colors. Once dry I used charcoal to draw a simple graphic version of Lady Liberty, smearing lightly with my finger. I highlighted these lines with white oil pastel. I added a few extra "scribbles" of color to mimic the energy of Max's work.

  1. These turned out beautiful.
    I love the combination of colors.

  2. these are AWESOME, Mary! I LOVE love love the colors and the black charcoal together! Beautiful!

  3. OOH! they look so good! Thanks for sharing! I think next time we'll use pastel, too :)


Content-Type: application/http; msgtype=response

WARC-Type: response

text:
Social Media

Daily Deals Funding, Yahoo's $610 Million Ruling, Instagram for Android: Today's Top Stories

Welcome to this morning's edition of “First To Know,” a series in which we keep you in the know on what’s happening in the digital world. We’re keeping our eyes on six particular stories of interest today.

Daily Deal Sites Raise Millions

The daily deals space is heating up again after Groupon's IPO: LivingSocial, and Plum District have raised between $20 million and $176 million apiece this week.

Yahoo Wins $610 Judgment in Lottery Scam

A judge granted Yahoo $610 million in a court ruling against groups that tried to scam people into handing over their passwords and other sensitive personal data.

House Passes Bill That Will Let Netflix Use Facebook's Open Graph

The U.S. House of Representatives has passed an outdated bill that prevents companies from sharing an individual’s movie-rental history, which would enable companies like Netflix to take advantage of social sharing features such as Facebook's open graph.

Alibaba Seeking $40 Million to Buy Back Stake

Chinese ecommerce giant Alibaba is seeking $40 million in debt financing to buy back Yahoo's 40% stake in the company.

Instagram Coming to Android

Popular iPhone photography app Instagram is finally developing an Android app.

Judge Hits Blogger With $2.5 Million Fine for Not Being a Journalist

In a case that’s sending a frightening message to the blogging community, a U.S. District Court judge ruled that a blogger must pay $2.5 million to an investment firm she wrote about — because she isn’t a real journalist.

Further News

Image courtesy of iStockphoto, DNY59

Load Comments

What's New

What's Rising

What's Hot

warcinfo:
robots: classic
hostname: ip-10-171-96-226.ec2.internal
software: Nutch 1.6 (CC)/CC WarcExport 1.0
isPartOf: CC-MAIN-2015-35
operator: CommonCrawl Admin
description: Wide crawl of the web for August 2015
publisher: CommonCrawl
format: WARC File Format 1.0
conformsTo: http://bibnum.bnf.fr/WARC/WARC_ISO_28500_version1_latestdraft.pdf


text:
Knotch Lets You Express Your Opinion in Full Color

Chances are you've given something you've enjoyed online a thumbs-up before, or rated your latest purchase or Netflix movie a 3 out of 5 stars, but have you ever expressed your opinion in color?

Knotch is an app that offers a different take on expressing your opinion. It allows users to express their opinions with colors, labeling everything from the latest smartphone to a television show in various hues that correspond to their feelings about that product or service. The app launched late last year, and updated to version 1.2 on Thursday.

“The idea initially was just to find a way to quantify opinions,” Knotch co-founder Anda Gansca told Mashable. “We were super-fascinated by how ads are targeted toward us.”

Knotch created a focus group to determine how people like to express their opinions, and in the end, they most liked the idea of expressing opinions in color form, taking a virtual temperature of how they feel about a particular topic.

With the app, you can type in a topic — or select from what’s currently being discussed on the platform — and weigh in on whether you feel hot or cold about it, selecting from a sliding scale of colors. Bright red indicates that a topic is something you enjoy, and dark blue means you feel "cold" about something, or don't like it.

More than just a color rating, you can also add a short comment to your posting to explain why you may have rated something the way you did.

“We wanted to find a way to bring the end user into the process of data collection and analysis in a way that’s incredibly public and empowering for the user,” says Gansca.

While the app lets you post your opinions — called knotches — on Facebook, most users choose to keep that information private in Knotch. Gansca says 80% of Knotch users never post to Facebook, instead building relationships with people in the app based on their similar opinions. 82% of users have opted to follow someone on the service outside of the social graph, forming a relationship based on their opinions.

Knotches biggest group of users — males between 24 and 35 — actually use the service like they would Reddit, spending more than 30 minutes a day in the app each day, knotching on different topics and commenting on others' posts.

With version 1.2, Knotch added Twitter integration for sharing knotches in tweets, and the ability to share your favorite knotches on Facebook. Discovery capabilities are also improved in the app, with a new Search routine for finding knotches on particular topics. You can also browse other knotches by topic — so you can find someone who loves your favorite band as much as you do — and a new Stream function lets you take a quick look at what’s currently going on globally on Knotch.

Knotch has also started rewarding its most active users. Giving them “glory” in topics they're passionate about, ultimately pulling out thought leaders on particular topics.

The Knotch team graduated from StartX, the Stanford incubator.

Knotch is available now from the Apple App Store.

What do you think about Knotch? Let us know your thoughts in the comments.

Thumbnail Image by Mashable, Emily Price; Screenshots courtesy of Knotch

Load Comments

What's New

What's Rising

What's Hot


text:
The Secret Of Monkey Island

5 Ways to Celebrate "Talk Like a Pirate Day" Online in 2010

Arrr! Ye Mashable readers know today be Talk Like a Pirate Day 2010, but were ye landlubbers aware that there be more ways t' celebrate online than there are bottles o' rum in ol' Jack Sparrow's locker? Thar be danger in the social media seas — sour ...

10 Classic PC Games That Found New Life on the iPhone

Apple's iPhone may mark a second golden age of computer gaming for a lot of reasons. The open marketplace means anyone can create a game; from major companies to just a couple of talented people in a garage. The result is a new boom in innovation, th...


WARC-Truncated: length

text:
Return to sender

Attorneys who have sent correspondence to the Suffolk County Probate & Family Court have been complaining that their mail has been returned. It turns out that the court stopped using its P.O. box last November. But Acting Register Sandra Giovannucci said the change was made after the 2013 Lawyers Diary and Manual had been printed, thus ...

Leave a Reply

Scroll To Top


text:
Motor vehicles – License plate cover

Where a judge allowed a defendant’s motion to suppress evidence found by a state trooper during a traffic stop, the suppression order must be affirmed because a tinted plastic cover over the vehicle’s rear license plate did not justify the stop. “At issue is a State trooper's decision, based on G.L.c. 90, §6, to stop a vehicle ...

Leave a Reply

Scroll To Top


text:
Author Archives: Kimberly Atkins

Feed Subscription

The pitfalls of ‘portability’

The new estate tax “portability” provision, which allows surviving spouses to benefit from the unused exclusion from their spouse’s estate, was a welcome part of the federal tax law enacted late last year for estate planning attorneys and their clients.

Read More »
Scroll To Top


text:
Land Court

Feed Subscription

Real property – Subdivision

Where the Edgartown planning board decided to deny ANR endorsement of the plaintiff’s plan to divide his lot in the Sweetened Water Farm subdivision, the board did not misapply the subdivision’s rules and regulations.

Read More »

Real property – Way – Easement rights

Where the plaintiff owners of land in Dennis have requested a judicial determination of the defendants’ rights in a 20-foot-wide way providing access to Cape Cod Bay, 28 of the defendants enjoy a right to use the way, while others do not.

Read More »

Real property – Taking – Easement

Where plaintiff landowners have requested a judgment declaring that a 1962 taking by the town of Lexington did not include their land, a judgment in favor of the plaintiffs is warranted based on the issues of notice and intent.

Read More »

Real property – Buffer zone – Time limit

Where a defendant has argued that the plaintiff neighbors are time-barred from enforcing a deed restriction that created a 40-foot-wide buffer zone that was “intended to be forever kept open and unimproved,” the restriction is unlimited as to time but the improvements made by the defendant within the buffer zone are not the sort that are encompassed by the G.L.c. 184, §23A six-year limitations period.

Read More »

Real property – Waste – Life tenants

Where two defendant holders of a life estate have not paid their share of property-related expenses to the plaintiff, their sister, who is the owner of the remainder, the defendants must either renounce their life tenancy or reimburse the plaintiff for her expenses.

Read More »

Real property – Condominium

Where plaintiff condominium trust has moved for summary judgment in a dispute over property which is part of phased condominium land, the motion must be denied based on language in a 2004 settlement agreement and an “Instrument of Revival and Grant.”

Read More »
Scroll To Top


text:
Official Suppliers of Better Turns

We are an international organization of top independent ski and snowboard shops that specialize in custom bootfitting and place a premium on excellent customer service.

Innovative Product.  Cutting Edge design!

We Manufacture a complete line of insoles and fit aid products.  Our products deliver superior quality , supreme comfort and uncompromising performance.

Global Training Centers for footwear fitters

Learn fit secrets from the top industry pros on state-on-the-art equipment at our training centers in the US, Austria and Australia.


text:
4/17/11 Rangers at Yankees: New Yankee Stadium

Ah Sunday Night Baseball. The lights, the people, and the only game going on at that time. The ballhawk’s nightmare. First, there is the normal weekend crowd. Then, there is the Sunday Night Baseball crowd. Then, there’s the fact it is a Yankee game.

So a ballhawk like myself would have to get there extra early right? Well, I didn’t. I set myself up to get to the ballpark at 5:30 but because of the MTA’s modified 1 train schedule on weekends which has it not working from my station to the stop before Yankee Stadium. In a nutshell, I got to the ballpark at 6:24, my ticket blew away into the street as I was told to take out my phone. Not a good day to that point.

Sorry for the lack of pictures during BP itself but I was in a frenzy trying to get a ball considering all the previously mention factors but I did manage to get a picture as I entered:
its grandness 41711.JPG

That managed to calm me down some but don’t let it fool you, the left field stands were absolutely crammed considering how big they are. Actually, I shouldn’t say that. There were more competent people than I expected. In Citi Field on the weekends it is actually easier to snag baseballs because everyone crams in the front row and there are no ballhawks to be found. Here, you had Zack Hample, two other people who looked like they knew what they were doing and a whole mess of roadblocks in the guise of fans.

From this, it is no surprise that my first ball was thrown. Seeing little hope for Home Run snags, I headed over to foul ground to get a ball from the pitchers warming up. I initially lined up behind Neftali Feliz because of the Spanish factor but moved onto new pairs as the former pairs didn’t throw me a ball.

Finally, Mark Lowe saw my Ranger’s gear and tossed me his warm-up ball for Ball #1 on the day. I was relieved. It seemd like the perfect set of circumstances to get shutout.

I then moved back into fair territory and as this happened a ball came to Neftali Feliz. He threw it back but I managed to get in the words, “la proxima, por favor?” which translates to, “the next one please”. Within five minutes another ball came to him and he lofted it to me for Ball #2 on the day. I was ecstatic. I had just matched my season average in Yankee Stadium on the weekend.

After this, I moved over to pull city in left field expecting security to come down any minute asking for people’s tickets. That’s when a person I identified as Adrian Beltre:
Beltre 41711.JPG

hit a Home Run a section to my left. I got there with plenty of time but from the corner of my eye it looked like someone was camped under the ball. I didn’t want to reach in front of the person so I backed off and prepared for a ricochet. But I then saw the person back off as they seemed to be afraid of the ball. I quickly put out my glove and snow-coned the ball. For those of you just following, this is a big deal because I am HORRENDOUS at tracking fly balls and catching them in the seats.

I moved back to my spot and after a whole five minutes Matt Harrison tossed me a ball for #4 on the day (being the Rangers fan that I am). This was by far my most efficient day snagging five balls in fifty minutes of Bp.
That’s right, five. First, I did a manuvering job to not get kicked out by security. Ironically after this message aired on the big screen:
Oh, Security 41711.JPG
It is under the red arrow and the ironic part says says, “Guests will be treated in a consistent professsional and courteous manner by Yankee Stadium Staff an the Yankees ask for the same in return.” Ha!
Then, Bp ended:
Bp ending 41711.JPG
I stayed by the Ranger’s bullpen as Alexei Ogando warmed up. As he played catch the bullpen catcher, Josh Frasier, went into the bullpen and tossed the hit balls to the bleacher creatures. After being encouraged, he tossed “one up to a Rangers fan”.
I knew that I wasn’t going to be getting anything else until gametime So, I took my spot in foul territory and look who showed up right in front of my eyes:

Kruk walking 41711.JPG

Under the red arrow would be ESPN’s John Kruk picking his favorite seat of Yankee Stadium as he will every Sunday Night Baseball.

Here he is doing the actual segment:

Best Seat 41711.JPG

The listed attendance was 40, 811 and for once it felt like that. This picture is from before the game and it is still pretty full:

Thumbnail image for attendance 41711.JPG

The snagging room was as good as it ever will be because of the brief spurts of rain/lightning.

The view to my left:

To the left 41711.JPG

The view to my right:

To the right 417111.JPG

Sadly, the only ball that came within 100 ft of me went into the second level just as I thought I had it caught.

The game as all I have gone to at Yankee Stadium was indeed exciting. The Yankees won again by scoring a run in the eighth off of Arthur Rhodes.

Rangers 5 Yankees 6

As I left, I saw the set atop a parking garage of Baseball Tonight:

BT set 41711.JPG

  • 5 Balls at this game

Balls 41711.JPG

Numbers 68-72
Sweet Spots 41711.JPG
  • 11 Balls in 4 games so far this season= an average of 2.75 Balls Per Game
  • 29 straight games with at least 1 ball
  • 6 straight at Yankee Stadium (just to clarify I only got one ball at the old Yankee Stadium waaay before this blog so Yankee Stadium refers to the new one)
  • Competition Factor 40811 fans* 5 balls= 204,055
  • Time at game 6:24-11:14= 4 hours and 50 minutes

Hello, observers of baseball. Let me know what you though of the entry no matter what you thought of it. Additonally, let me know if you'd like to see me do anything in specific or what I'm doing right or wrong when I do write entries.

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Get every new post delivered to your Inbox.

Join 478 other followers

%d bloggers like this:


text:
Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let's say I take a ton of measurements of the diameter of a marble with a 0.001" resolution micrometer, and I calculate the standard deviation of the sample set. Would the standard deviation be a better measure of uncertainty than the resolution of the instrument?

share|improve this question
Better for what purpose? –  Qiaochu Yuan Sep 17 '12 at 3:05

2 Answers 2

up vote 1 down vote accepted

The resolution on the meter only tells you that you can't clearly resolve measurements observed between tick marks. It does not tell you how consistently the results are recorded. Repeated measurement of an object of a known length will directly measure the recording error variability through the estimated standard deviation. This will not be exact. It is subject to sampling variability (i.e. it is a sample estimate of a population parameter).

The standard deviation is a measure of the accuracy which includes meaasurement precision and bias in recording whereas instrument resolution only measures precision. So it can understate accuracy. The sample variance will be statistically an unbiased estimate of variance which does measure accuracy.

share|improve this answer

The standard deviation tells you how precise the measurement is. It tells you what the energy of the measurement error is.

If you have actual bounds on the magnitude of the measurement error, you can use interval arithmetic. I love interval arithmetic, but it can produce over-pessimistic bounds.

share|improve this answer

Your Answer

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.


text:
Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\ I_n = \int \cos^n \theta d\theta$

Find a recursive equation to express $\ I_n $ in terms of $\ I_n-_2 $. Find $\ I_6 $ and $\ I_7 $ as well.

share|improve this question
en.wikipedia.org/wiki/… –  TenaliRaman Nov 23 '12 at 17:34

2 Answers 2

Hint: $\cos^n(x) = \cos(x)\cos^{n-1}(x)$ and try integrating by parts. Also: remember that $\sin^2(x) + \cos^2(x) = 1$

share|improve this answer

We show how to avoid explicit integration by parts. Guess that $f(x)=\cos^{n-1} x\sin x$ works. Check by differentiating. We get $$f'(x)= \cos^n x-(n-1)\sin^2 x\,\cos^{n-2} x.$$ The first part looks good. For the second part, write $\sin^2 x=1-\cos^2 x$.

We get $$f'(x)=n\cos^n x-(n-1)\cos^{n-2} x.$$ It follows that $$n\int \cos^n x\,dx=\cos^{n-1}x\, \sin x +(n-1)\int \cos^{n-2} x\,dx.$$

share|improve this answer

Your Answer

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.


text:
[Physics FAQ] - [Copyright]

Original by Michael Weiss and John Baez.

Is Energy Conserved in General Relativity?

In special cases, yes.  In general — it depends on what you mean by "energy", and what you mean by "conserved".

In flat spacetime (the backdrop for special relativity) you can phrase energy conservation in two ways: as a differential equation, or as an equation involving integrals (gory details below).  The two formulations are mathematically equivalent.  But when you try to generalize this to curved spacetimes (the arena for general relativity) this equivalence breaks down.  The differential form extends with nary a hiccup; not so the integral form.

The differential form says, loosely speaking, that no energy is created in any infinitesimal piece of spacetime.  The integral form says the same for a finite-sized piece.  (This may remind you of the "divergence" and "flux" forms of Gauss's law in electrostatics, or the equation of continuity in fluid dynamics.  Hold on to that thought!)

An infinitesimal piece of spacetime "looks flat", while the effects of curvature become evident in a finite piece.  (The same holds for curved surfaces in space, of course).  GR relates curvature to gravity.  Now, even in newtonian physics, you must include gravitational potential energy to get energy conservation.  And GR introduces the new phenomenon of gravitational waves; perhaps these carry energy as well?  Perhaps we need to include gravitational energy in some fashion, to arrive at a law of energy conservation for finite pieces of spacetime?

Casting about for a mathematical expression of these ideas, physicists came up with something called an energy pseudo-tensor.  (In fact, several of 'em!) Now, GR takes pride in treating all coordinate systems equally.  Mathematicians invented tensors precisely to meet this sort of demand — if a tensor equation holds in one coordinate system, it holds in all.  Pseudo-tensors are not tensors (surprise!), and this alone raises eyebrows in some circles.  In GR, one must always guard against mistaking artifacts of a particular coordinate system for real physical effects.  (See the FAQ entry on black holes for some examples.)

These pseudo-tensors have some rather strange properties.  If you choose the "wrong" coordinates, they are non-zero even in flat empty spacetime.  By another choice of coordinates, they can be made zero at any chosen point, even in a spacetime full of gravitational radiation.  For these reasons, most physicists who work in general relativity do not believe the pseudo-tensors give a good local definition of energy density, although their integrals are sometimes useful as a measure of total energy.

One other complaint about the pseudo-tensors deserves mention.  Einstein argued that all energy has mass, and all mass acts gravitationally.  Does "gravitational energy" itself act as a source of gravity?  Now, the Einstein field equations are

            Gmu,nu = 8pi Tmu,nu
Here Gmu,nu is the Einstein curvature tensor, which encodes information about the curvature of spacetime, and Tmu,nu is the so-called stress-energy tensor, which we will meet again below.  Tmu,nu represents the energy due to matter and electromagnetic fields, but includes NO contribution from "gravitational energy".  So one can argue that "gravitational energy" does NOT act as a source of gravity.  On the other hand, the Einstein field equations are non-linear; this implies that gravitational waves interact with each other (unlike light waves in Maxwell's (linear) theory).  So one can argue that "gravitational energy" IS a source of gravity.

In certain special cases, energy conservation works out with fewer caveats.  The two main examples are static spacetimes and asymptotically flat spacetimes.

Let's look at four examples before plunging deeper into the mathematics.  Three examples involve redshift; the other, gravitational radiation.

Very fast objects emitting light

According to special relativity, you will see light coming from a receding object as redshifted.  So if you, and someone moving with the source, both measure the light's energy, you'll get different answers.  Note that this has nothing to do with energy conservation per se.  Even in newtonian physics, kinetic energy (mv^2/2) depends on the choice of reference frame.  However, relativity serves up a new twist.  In newtonian physics, energy conservation and momentum conservation are two separate laws.  Special relativity welds them into one law, the conservation of the energy-momentum 4-vector.  To learn the whole scoop on 4-vectors, read a text on SR, for example Taylor and Wheeler (see refs.) For our purposes, it's enough to remark that 4-vectors are vectors in spacetime, which most people privately picture just like ordinary vectors (unless they have very active imaginations).

Very massive objects emitting light

Light from the Sun appears redshifted to an Earth bound astronomer.  In quasi-newtonian terms, we might say that light loses kinetic energy as it climbs out of the gravitational well of the Sun, but gains potential energy.  General relativity looks at it differently.  In GR, gravity is described not by a "potential" but by the "metric" of spacetime.  But "no problem", as the saying goes.  The Schwarzschild metric describes spacetime around a massive object, if the object is spherically symmetrical, uncharged, and "alone in the universe".  The Schwarzschild metric is both static and asymptotically flat, and energy conservation holds without major pitfalls.  For more details, consult MTW, chapter 25.

Gravitational waves

A binary pulsar emits gravitational waves, according to GR, and one expects (innocent word!) that these waves will carry away energy.  So its orbital period should change.  Einstein derived a formula for the rate of change (known as the quadrapole formula), and in the centenary of Einstein's birth, Russell Hulse and Joseph Taylor reported that the binary pulsar PSR1913+16 bore out Einstein's predictions within a few percent.  Hulse and Taylor were awarded the Nobel prize in 1993.

Despite this success, Einstein's formula remained controversial for many years, partly because of the subtleties surrounding energy conservation in GR.  The need to understand this situation better has kept GR theoreticians busy over the last few years.  Einstein's formula now seems well-established, both theoretically and observationally.

Expansion of the universe leading to cosmological redshift

The Cosmic Background Radiation (CBR) has red-shifted over billions of years.  Each photon gets redder and redder.  What happens to this energy?  Cosmologists model the expanding universe with Friedmann-Robertson-Walker (FRW) spacetimes.  (The familiar "expanding balloon speckled with galaxies" belongs to this class of models.) The FRW spacetimes are neither static nor asymptotically flat.  Those who harbor no qualms about pseudo-tensors will say that radiant energy becomes gravitational energy.  Others will say that the energy is simply lost.

It's time to look at mathematical fine points.  There are many to choose from!  The definition of asymptotically flat, for example, calls for some care (see Stewart); one worries about "boundary conditions at infinity".  (In fact, both spatial infinity and "null infinity" clamor for attention — leading to different kinds of total energy.) The static case has close connections with Noether's theorem (see Goldstein or Arnold).  If the catch-phrase "time translation symmetry implies conservation of energy" rings a bell (perhaps from quantum mechanics), then you're on the right track.  (Check out "Killing vector" in the index of MTW, Wald, or Sachs and Wu.)

But two issues call for more discussion.  Why does the equivalence between the two forms of energy conservation break down?  How do the pseudo-tensors slide around this difficulty?

We've seen already that we should be talking about the energy-momentum 4-vector, not just its time-like component (the energy).  Let's consider first the case of flat Minkowski spacetime.  Recall that the notion of "inertial frame" corresponds to a special kind of coordinate system (minkowskian coordinates).

Pick an inertial reference frame.  Pick a volume V in this frame, and pick two times t=t0 and t=t1.  One formulation of energy-momentum conservation says that the energy-momentum inside V changes only because of energy-momentum flowing across the boundary surface (call it S).  It is "conceptually difficult, mathematically easy" to define a quantity T so that the captions on the Equation 1 (below) are correct.  (The quoted phrase comes from Sachs and Wu.)

  Equation 1:  (valid in flat Minkowski spacetime, when Minkowskian
                coordinates are used)

       /                  /                    /
       |                  |                    |
       | T dV     -       | T dV       =       | T dt dS
       /                  /                    /
     V,t=t0             V,t=t1                 t=t0

   p contained       p contained            p flowing out through
   in volume V    -  in volume V       =    boundary S of V
   at time t0        at time t1             during t=t0 to t=t1

   (Note: p = energy-momentum 4-vector)
T is called the stress-energy tensor.  You don't need to know what that means! —just that you can integrate T, as shown, to get 4-vectors.  Equation 1 may remind you of Gauss's theorem, which deals with flux across a boundary.  If you look at Equation 1 in the right 4-dimensional frame of mind, you'll discover it really says that the flux across the boundary of a certain 4-dimensional hypervolume is zero.  (The hypervolume is swept out by V during the interval t=t0 to t=t1.)  MTW, chapter 7, explains this with pictures galore.  (See also Wheeler.)

A 4-dimensional analogue to Gauss's theorem shows that Equation 1 is equivalent to:

  Equation 2:  (valid in flat Minkowski spacetime, with Minkowskian

       coord_div(T) = sum_mu (partial T/partial x_mu) = 0
We write "coord_div" for the divergence, for we will meet another divergence in a moment.  Proof?  Quite similar to Gauss's theorem: if the divergence is zero throughout the hypervolume, then the flux across the boundary must also be zero.  On the other hand, the flux out of an infinitesimally small hypervolume turns out to be the divergence times the measure of the hypervolume.

Pass now to the general case of any spacetime satisfying Einstein's field equation.  It is easy to generalize the differential form of energy-momentum conservation, Equation 2:

  Equation 3:  (valid in any GR spacetime)

        covariant_div(T) = sum_mu nabla_mu(T) = 0

                    (where nabla_mu = covariant derivative)
(Side comment: Equation 3 is the correct generalization of Equation 1 for SR when non-minkowskian coordinates are used.)

GR relies heavily on the covariant derivative, because the covariant derivative of a tensor is a tensor, and as we've seen, GR loves tensors.  Equation 3 follows from Einstein's field equation (because something called Bianchi's identity says that covariant_div(G)=0).  But Equation 3 is no longer equivalent to Equation 1!

Why not?  Well, the familiar form of Gauss's theorem (from electrostatics) holds for any spacetime, because essentially you are summing fluxes over a partition of the volume into infinitesimally small pieces.  The sum over the faces of one infinitesimal piece is a divergence.  But the total contribution from an interior face is zero, since what flows out of one piece flows into its neighbor.  So the integral of the divergence over the volume equals the flux through the boundary.  "QED".

But for the equivalence of Equations 1 and 3, we would need an extension of Gauss's theorem.  Now the flux through a face is not a scalar, but a vector (the flux of energy-momentum through the face).  The argument just sketched involves adding these vectors, which are defined at different points in spacetime.  Such "remote vector comparison" runs into trouble precisely for curved spacetimes.

The mathematician Levi-Civita invented the standard solution to this problem, and dubbed it "parallel transport".  It's easy to picture parallel transport: just move the vector along a path, keeping its direction "as constant as possible".  (Naturally, some non-trivial mathematics lurks behind the phrase in quotation marks.  But even pop-science expositions of GR do a good job explaining parallel transport.) The parallel transport of a vector depends on the transportation path; for the canonical example, imagine parallel transporting a vector on a sphere.  But parallel transportation over an "infinitesimal distance" suffers no such ambiguity.  (It's not hard to see the connection with curvature.)

To compute a divergence, we need to compare quantities (here vectors) on opposite faces.  Using parallel transport for this leads to the covariant divergence.  This is well-defined, because we're dealing with an infinitesimal hypervolume.  But to add up fluxes all over a finite-sized hypervolume (as in the contemplated extension of Gauss's theorem) runs smack into the dependence on transportation path.  So the flux integral is not well-defined, and we have no analogue for Gauss's theorem.

One way to get round this is to pick one coordinate system, and transport vectors so their components stay constant.  Partial derivatives replace covariant derivatives, and Gauss's theorem is restored.  The energy pseudo-tensors take this approach (at least some of them do).  If you can mangle Equation 3 (covariant_div(T) = 0) into the form:

       coord_div(Theta) = 0
then you can get an "energy conservation law" in integral form.  Einstein was the first to do this; Dirac, Landau and Lifshitz, and Weinberg all came up with variations on this theme.  We've said enough already on the pros and cons of this approach.

We will not delve into definitions of energy in general relativity such as the hamiltonian (amusingly, the energy of a closed universe always works out to zero according to this definition), various kinds of energy one hopes to obtain by "deparametrizing" Einstein's equations, or "quasilocal energy".  There's quite a bit to say about this sort of thing!  Indeed, the issue of energy in general relativity has a lot to do with the notorious "problem of time" in quantum gravity. . . but that's another can of worms.

References (vaguely in order of difficulty):


text:
Home > data science, rant > Fighting the information war (but only on behalf of rich people)

Fighting the information war (but only on behalf of rich people)

December 11, 2012

There’s an information war out there which we have to be prepared for. Actually there a few of them.

And according to this New York Times piece, there’s now a way to fight against the machine, for a fee. Companies like Reputation.com will try to scour the web and remove data you don’t want floating around about you, and when that’s impossible they’ll flood the web with other good data to balance out the bad stuff.

At least that’s what I’m assuming they do, because they of course don’t really explain their techniques. And that’s the other information war, where they scare rich people with technical sounding jargon and tell them unlikely stories to get their money.

I’m not claiming predatory information-gatherers aren’t out there. But this is the wrong way to deal with it.

First of all, most of the data out there systematically being used for nefarious purposes, at least in this country, is used against the poor, denying them reasonable terms on their loans and other services. So the idea that people will need to pay for a service to protect their information is weird. It’s like saying the air quality is bad for poor people, so let’s charge rich people for better air.

So what kind of help is Reputation.com actually providing? Here’s my best guess.

First it targets people to get overly scared in the spirit of this recent BusinessWeek article, which explains that cosmetic companies have gone to China and started a campaign to convince Chinese women they are too hairy so they’ll start buying products to remove hair. From that article, which is guaranteed to make you understand something about American beauty culture too:

Despite such plays on women’s fears of embarrassment, Reckitt Benckiser’s Sehgal says that Chinese women are too “independent-minded” to be coaxed into using a product they don’t really need. Others aren’t so sure. Veet’s Chinese marketing “plays a role that is very similar to that of the apple in the Bible,” says Benjamin Voyer, a social psychologist and assistant professor of marketing at ESCP Europe business school. “It creates an awareness, which subsequently creates a feeling of shame and need.”

Second, Reputation.com gets their clients off nuisance lists, like the modern version of a do-not-call program (which, importantly, is run by the government). This is probably equivalent to setting up a bunch of email filters and clearing their cookies every now and then, but they can’t tell their clients that.

Finally, for those rich people who are also super vain, they will try to do things like replace the unflattering photos of them that come up in a google image search with better-looking ones they choose. Things like that, image issues.

I just want to point out one more salient fact about Reputation.com. It’s not just in their interest to scare-monger, it’s actually in their interest to make the data warehouses more complete (they have themselves amassed an enormous database on people), and to have people who don’t pay for their services actually need their services more. They could well create a problem to produce a market for their product.

What drives me nuts about this is how elitist it is.

There are very real problems in the information-gathering space, and we need to address them, but one of the most important issues is that the very people who can’t afford to pay for their reputation to be kept clean are the real victims of the system.

There is literally nobody who will make good money off of actually solving this problem: I challenge any libertarian to explain how the free market will address this. It has to be addressed through policy, and specifically through legislating what can and cannot be done with personal data.

Probably the worst part is that, through using the services from companies Reputation.com and because of the nature of the personalization of internet usage, the very legislators who need to act on behalf of their most vulnerable citizens won’t even see the problem since they don’t share it.

Categories: data science, rant
  1. JSE
    December 11, 2012 at 8:36 am

    “There is literally nobody who will make good money off of actually solving this problem: I challenge any libertarian to explain how the free market will address this.”

    How confident are you that libertarians think there’s a problem?

    • December 11, 2012 at 10:35 am

      Pretty confident. Most libertarians I talk to agree that there shouldn’t be a systematic underclass. But they argue that market forces will keep it from happening. But I could be wrong.

      • libertarian by default
        December 11, 2012 at 1:53 pm

        How does the information in question become searchable by nefarious agents in the first place? I believe the primary market mechanism to “fix” this is for consumers to favor competitors that don’t leak the information and/or pay a premium for an explicit agreement re: privacy. The supermarket discount cards are a good example of a market solution: lower prices in exchange for letting them track your purchases.

        • December 11, 2012 at 1:57 pm

          Yes but even if we knew exactly who sells our info and to whom, which we don’t right now because a lack of transparency, it’s exactly the people who can’t afford to pass up a discount who should be protecting their data. Again you’re asking for people to pay for privacy.

  2. JSE
    December 11, 2012 at 8:39 am

    “So the idea that people will need to pay for a service to protect their information is weird. It’s like saying the air quality is bad for poor people, so let’s charge rich people for better air.” Some people say the reason that clean air regulation has been so successful is related to the fact that it’s hard to keep rich and poor people from breathing the same air.

    Maybe better is to say “It’s like saying that terrible public schools are bad for poor people, so let’s charge rich people for better schools.” Which may or may not be weird, but in many parts of the country it’s the system we actually have.

    • December 11, 2012 at 9:30 am

      yes, that’s even better, thanks. But the problem with both examples is how easy it is to see if air is bad or if education is bad, but how individual experiences on the web are harder to measure.

      • JSE
        December 11, 2012 at 10:20 am

        Actually, I think the air analogy is a good one in this respect — SOME kinds of air pollution are easy to see, but others are not, and the invisible ones are dangerous, too. Some invasions of privacy are visible to the victim, too, but not all.

  3. Mo
    December 11, 2012 at 10:57 am

    I challenge any Statist to explain how “policy” and “legislation” will fullfil a supposed need of ignorants that intentionally and voluntarily uploaded data into the public domain and wish it would not be used by third parties for forming an opinion about those individuals.

    For example,
    1 – A hypothetical “Cathy” uses {Twitter/Facebook/Blogs/pick your favorite} to tell the world about her appetite for gallons of booze / how much she’s losing on poker and the stock market / how she broke her collarbone for the nth time while skateboarding.
    2. A potential employer / loan shark / health insurer uses above data from the public domain to form an opinion about our hypothetical “Cathy”.
    3. The potential service provider refuses service based on that opinion.

    So what’s the proposed “policy”? Unleash violence against any party that tries to form an opinion about someone else based on public data? Unleash violence against any service provider that refuses to provide a service based on public data about the service requester?

    Oh wait, I think I got it: let’s create a new government agency called the “Ministry of Truth” for “fixing” everyone’s online reputation, “for free” of course…

    Sry for being in a sarcastic mood, usually appreciate this blog

    • December 11, 2012 at 11:02 am

      Good question. I don’t think we will ever be able to keep individuals from googling people’s names and seeing what shows up. And even teenagers nowadays realize that what they put up on the web is there for good.

      What I’m more objecting to is systematized and statistically-driven classism and racism. If I grew up poor, and minority, I should still be able to apply for and receive a loan based on my personal merits, not the statistical properties of people like me. But there’s no regulation for how online data (like zip code etc.) is used, and loans decisions are increasingly made based on those data and patterns.

      • libertarian by default
        December 11, 2012 at 2:59 pm

        I strongly disagree that we need regulation regarding “zipcode” in a loan decision.

        It seems like a creditor is naturally incentivized to figure out the best possible variables to predict loan default (all regulation can possibly do is distort this pretty natural problem of wanting to loan money to those most likely to pay it back). If we suspect e.g. racism plays a role here, then presumably BigData will only help that problem by eventually showing that the “race” variable is marginally useless given a bunch of other factors that it’s (in our flawed world) unfortunately correlated with (also probably removing more and more of the “human judgement” in a loan approval).

        To the extent some banks use things like “zipcode” as a filter (perhaps because they learn it’s too expensive to dig deeper, which isn’t so much evil as just pragmatic) it should open opportunities for a “leaner, smarter” bank to find an underserviced population.

        Finally, if there are variables considered unsavory in loan applications, nothing prevents a competitor from explicitly ignoring them and advertising as such. To the extent they were truly unsavory variables (i.e. not useful predictors given the rest of the available data) this is another avenue for market forces to solve the “zipcode” problem without government intervention.

      • Mo
        December 11, 2012 at 3:47 pm

        I Agree that being misclassified by a racist predictive model is not exactly something that you can be held accountable for as an individual (as opposed to the examples I provided in my previous comment). However, let’s zoom in on such a racist predictive model – which is either:

        1. Inaccurate because of its racist bias. In that case, there is a huge business opportunity for those who are able to produce a more accurate predictive model. There’s our free market solution.
        2. Reasonably(*) accurate, with no significant room for optimization. What do you propose to do for the few unlucky ones that could have received a loan based on personal merits? Does being unlucky give you (or any entity representing you) the right to force a third party into providing you the service you feel being entitled to?

        (*) “Reasonable”, as in providing the employer of the model a positive expected value on bets such as loans or insurances.

        • December 11, 2012 at 3:55 pm

          Or a third option. Namely that the model reinforces racist history and therefore effects it. It’s a feedback loop people. We should know that by now.

        • Mo
          December 12, 2012 at 11:16 am

          Talking about feedback loops, here’s another one: politicians have a huge incentive to maintain poverty. Would they be of any use in a prosperous society where most people can take care of themselves without needing any entitlements? As a politician, how will you get votes or find cannon fodder for overseas military adventures without a “poor class” in society?

          I have a strong gut feeling that the business of politics itself has a stronger weight in reinforcing poverty or racism than some biased predictive models used by insurers or loan sharks…

  4. December 11, 2012 at 7:42 pm

    I can’t solve the data problem, but this brings to mind the NoCoat Tongue Scraper story from Infinite Jest. It’s pretty much the Veet story.

    “It did what all ads are supposed to do: create an anxiety relievable by purchase.” -DFW

    • Mo
      December 12, 2012 at 11:54 am

      Marketeers abusing anxiety is just a symptom: the anxiety itself is the problem that should be solved in society.

  5. GD
    December 11, 2012 at 11:29 pm

    The solution is fairly straightforward and goes back how this problem started in the first place. It turns out that was way before the days of the Internet. The first landmark was using Social Security Numbers as unique financial identifiers (breaking the usual political promise that SSNs would never be used as a national identifier). The second was Congress’ invention of the credit reporting biz in the 1970s. Since then practically all info is bought and sold based on SSN’s and it is the reason that identity theft is a cottage industry. Even victims of identity theft are prevented from dealing effectively with that problem when they are given a time limited “credit freeze”.

    Allow people to be removed form the credit reporting data banks and you will have a whole new ball game. And don’t get me started with how crucial that is for getting loans and doing business. Believe me there are plenty of people out there who don’t care. I am one of them.

  6. Brian
    December 12, 2012 at 10:43 am

    For all the scenarios of ‘private’ information being compromised by individuals self-posting info on their own Facebook pages, the more interesting issues of privacy revolve around indirect sources of information, covert collection, and fusion of separate sourcs.

    If a family member or acquaintance stores your phone, email, home address, employers name, etc on their own Cloud service provider app, do you mind it getting out, or being sold to third parties? Do you feel you have any right to control or mitigate sharing of your private data?

    Suppose a storefront decides to use a wireless sniffing service to log the dates and times of cell phones in their proximity; which is an available service for marketing research. Do you mind having your movements made available? What if the establishment is an adult bookstore that is near, or on the path to, where you often eat lunch? Would you like your name and other data made available in a marketing list for likely adult store patrons? Such a list can also be filtered to generate pairs or groups of people that are often together, do you want you and your coworker to be in that list? Do you feel any need or desire to know that it is there, or to be removed from it?

    These are just a few very simple scenarios that can illustrate the legitimate questions regarding privacy that are very possible with technology and services available today. Society needs to work through these types of issues, so please try to focus the discussion on possible scenarios where legitimate differences of opinion exist rather than sticking with overly simplistic discussions of people being upset that their public presence was viewed improperly.

    • Mo
      December 12, 2012 at 11:52 am

      Good points, I generally agree with the suggestion that society nowadays cannot keep up with technological advances, it’s clear that many service providers have issues with data security and privacy. However, the “free-market solution” asked for is quite simple: nobody is forced to use the advanced technological devices that endanger one’s privacy.

      If one chooses to use such devices or services (e.g. Facebook or a modern cell phone), it’s one’s responsibility to understand the terms of use and the risks involved. Maybe we can even come to the conclusion that usage of modern data-driven (and data-spreading) devices and services together with a guarantee of privacy has a price, as compared to the existing “cheap” (or even “free”) versions.

      It all comes down to the matter of well-defined contracts between service providers and consumers. Only when a contract is breached, one can speak of a “crime” that gives the right to any authority to intervene in the situation.

  7. blur blaster
    December 13, 2012 at 5:36 am

    Wouldn’t it be ironic if they couldn’t clean their own reputation?

  8. Karen
    December 10, 2013 at 11:51 pm

    Don’t know if anyone will see this so long after the first appearance, but one use for this service is to make it hard for an ex-wife to track the father of her children for unpaid child support. My ex purged his presence off the Internet (whether himself, or with help–he’s a computer guru) to make locating him difficult. It took 6 years.

  1. December 29, 2012 at 7:49 am
  2. January 6, 2013 at 1:16 am
  3. January 21, 2013 at 6:48 am
  4. April 9, 2013 at 6:08 am
Comments are closed.

Get every new post delivered to your Inbox.

Join 3,605 other followers

%d bloggers like this:


text:
Drexel dragonThe Math ForumDonate to the Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Math Forum » Discussions » Software » comp.soft-sys.matlab

Topic: matrix from random selected data
Replies: 2   Last Post: Nov 7, 2013 6:08 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 2
Registered: 11/7/13
Re: matrix from random selected data
Posted: Nov 7, 2013 6:08 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

> % Its not clear from your description how the "new"
> % matrix is derived from the original 100X100 matrix.
> % Can you provide an example?

I mean may the original matrix is name A is 100x100 and some data qualify, so we can take it from the matrix...
but i need to know, neighbor data from the original. So I make matrix
and the new one is name A1,A2,A3...to An ....so all selected data have new matrix 3x3

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum 1994-2015. All Rights Reserved.


text:
Drexel dragonThe Math ForumDonate to the Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Math Forum » Discussions » Software » comp.soft-sys.matlab

Topic: Q: Double integration in Matlab 5.0
Replies: 1   Last Post: Jul 15, 1997 1:06 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
This is Anfield

Posts: 3
Registered: 12/7/04
Q: Double integration in Matlab 5.0
Posted: Jul 15, 1997 11:30 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

I was using the suggestion in Technical notes 1502
(dblquad.m and innerlp.m) to perform a double integration.
The method works in Matlab 4.2
But in Matlab 5.0, I got the following error message:

??? Undefined function or variable 'QUAD_RECURSION_COUNT'.

Error in ==> /software/matlab50/toolbox/matlab/funfun/quad.m
On line 79 ==> if ~isempty(QUAD_RECURSION_COUNT) & (QUAD_RECURSION_COUNT > 1)

Errorr in ==> dblquad.m
On line 14 ==> Q = quad('innerlp', outmin, outmax, tol, trace, intfcn, ...

Any ways to get around this problem please ?

Tony Chan

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum 1994-2015. All Rights Reserved.


text:
Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there a smooth compact complex curve that does not admit an injective holomorphic map to $\mathbb CP^2$ ? Let me stress, that the image of the curve in $\mathbb CP^2$ can have singularities.

I should probably add, that I see this question as a baby version of the following question of Fancesco Polizzi: Surfaces in $\mathbb{P}^3$ with isolated singularities

Remark. Any curve $x^n+y^n+z^n=0$ admits injective morphisms to $\mathbb CP^2$ of arbitrary high degree (the proof is given by Jeremy Blanc here: Injective morphism from an elliptic curve to $\mathbb CP^2$. )

share|improve this question
If the map is an closed embedding, then the curve is smooth in the plane. Writing $d$ its degree, the adjunction formula says that it has genus $(d-1)(d-2)/2$, so of course not all curves are possible. If the map is not an embedding, the image is singular, with only cuspidal singularities. There are thus more possibilities for the curve. In this case, I don't know if each curve is possible. –  Jérémy Blanc Feb 22 '13 at 18:28
@Jérémy: I assume that aglearner is allowing unibranch singularities. –  Daniel Litt Feb 22 '13 at 19:08
@Jérémy: why must the singularities be cuspidal? –  Jesus Martinez Garcia Feb 22 '13 at 19:11
@Daniel: for me a cusp is exactly a unibranch singularity, as for most people studying cuspidal curves. @Jesus: the injectivity implies that when you make a resolution of singularities, there should be only one point above each singular pt. This is equivalent to ask for a cusp (or unibranch singularity). –  Jérémy Blanc Feb 23 '13 at 9:56
Oh, OK. I thought cusp meant $y^2-x^3$ (never read it anywhere, just from conversations), but I realise that analytically that is also $y^m-x^n$ for $m,n>1$ , which I think is equivalent to what you say. Thanks a lot :) –  Jesus Martinez Garcia Feb 26 '13 at 13:16

3 Answers 3

Estimates coming from the Bogomolov-Miyaoka-Yau inequality yield:

Corollary. (to Theorem A in this paper of Sakai) There is no $M$ such that every curve admits an injective morphism to a curve in $\mathbb{P}^2$ whose points all have multiplicity at most $M$.

Proof. The geometric genus $g$ curves of degree $d$ move in a family of dimension $3d + g - 1$, so if these are to fill the moduli space we must have $2g \le 3d + 2$. On the other hand, Theorem A of this paper of Sakai gives after rearranging that for any genus $g$, degree $d$, unibranch plane curve, $$ 2g \ge \frac{d^2}{2m+1} + O(md) $$ where $m$ is the maximal multiplicity of a point on the curve. For any bounded $m$, these inequalities cannot both be satisfied for large $g$. $\square$

Remark: the argument also shows that the same statement holds restricted to hyperelliptic curves. This is not incompatible with Jeremy's answer below, which requires a cusp of multiplicity $2g$ to inject a hyperelliptic curve of genus $g$.

I leave the older argument below because it is a different and possibly still useful approach.

Theorem. There is no finite set $\Sigma$ of topological types of unibranch singularities such that every smooth curve admits an injective map to a curve in $\mathbb{P}^2$ with only singularities from $\Sigma$.

Proof. I will argue that curves with unibranch singularities of total cogenus $\delta$ occur in the linear system $|\mathcal{O}(d)|$ with codimension at least $\mathrm{min}(\kappa, 3d - 1 + \delta)$, where $\kappa$ is the codimension of the equiclassical locus in the versal deformation of the singularities; it is $2\delta$ in the case of curves with only cusps, and for any finite set of singularities is bounded below by $(1 + \epsilon)\delta$ for $0 < \epsilon \le 1$ some constant depending on the set of singularities. Let us first see why this implies the result.

Saying a geometric genus $g$ curve appears in degree $d$ in codimension $3d - 1 + \delta = 3d + (d-1)(d-2)/2 - g - 1$ means it comes in a $g + 1 + (d+1)(d+2)/2 - (d-1)(d-2)/2 -3d = g + 3$ dimensional family. On the other hand $\kappa$ is given by $\delta + r$, where $r$ is the total ramification divisor on the normalization of the curve (for the normalization map), i.e. the sum over the branches of the singularities of (multiplicity - 1). So in this case the curves come in a family of dimension $g + 4 + 3d - r$. I don't know what to do with this, but under the further assumption that the set of singularities from which we choose must be finite and so we get $\kappa = (1 + \epsilon) \delta$ as above (e.g., for only cusps, $\epsilon = 1$), then the argument in Will's answer above shows this can fill the moduli space for only finitely many $g$.

To get the advertised codimension estimate, let $C$ be such a curve of degree $d$. Let $\Lambda$ be the tangent space to the versal deformation of the singularities of $C$. Then there exists a linear subspace $L \subset \Lambda$ such that

  1. equigeneric ideal $\ge$ L $\ge$ equiclassical ideal
  2. aside from the above restriction, L may be taken so that the dimension of the equigeneric ideal mod L is up to $3d - 1$.
  3. the image of $T_C|\mathcal{O}(d)| \to \Lambda$ is transverse to $L$.

This is essentially classical, for the sort of argument to prove it you can see e.g. the statement of Lemma 8 and the proof of Lemma 10 in my paper with Steve. $\square$

It's not yet clear to me how to improve this to answer the OP, maybe the better sort of bounds of Greuel, Lossen, and Shustin will help? (Or maybe there's just some easy argument not involving deformation theory...)

share|improve this answer
Vivek, thanks a lot! It will take me time to absorb your proof. –  aglearner Feb 26 '13 at 9:48
I'll try and write about the deformation theory when I'm less lazy (& assuming no-one puts a proof of the whole thing...) –  Vivek Shende Feb 26 '13 at 23:10
Vivek, huge thanks for this update! Now I am completely convinced in your statement. Your answer is really nice (the only reason I have not yet accepted it is that it seems to me now that the original question is an open problem...) –  aglearner Mar 13 '13 at 21:39
Having asked around a bit, I also think it's open. The fact that hyperelliptic curves inject, but not with bounded multiplicity, suggests that giving an answer of 'yes' to your original question requires a subtler analysis that sees the difference between 2g and 3g which sounds discouraging. I haven't given up yet though! –  Vivek Shende Mar 14 '13 at 12:21

The moduli space of curves of degree $d$ in $\mathbb P^2$ has dimension $\left( \begin{array}{c} d + 2 \\ 2 \end{array}\right)$. The most typical type of singularity is a simple cusp. Having a cusp at a given point is a codimension $4$ condition, so having $k$ cusps anywhere is codimension $2k$. This gives a moduli space of dimension $\left( \begin{array}{c} d + 2 \\ 2 \end{array}\right) -2k$ of objects of geometric genus $\left( \begin{array}{c} d -1 \\ 2 \end{array}\right)-k$, since each cusp reduces the geometric genus, relative to the arithemetic genus, by $1$.

Setting $k=\left( \begin{array}{c} d -1 \\ 2 \end{array}\right)-g$, if we wish these curves to fill up the moduli space of curves of genus $g$, $d$ and $g$ must satisfy the inequality:

$\left( \begin{array}{c} d + 2 \\ 2 \end{array}\right) +2g -2 \left( \begin{array}{c} d -1 \\ 2 \end{array}\right) \geq 3g-3$

$g \leq 3 + \left( \begin{array}{c} d + 2 \\ 2 \end{array}\right) - 2\left( \begin{array}{c} d -1 \\ 2 \end{array}\right)$

which is satisfiable for only finitely many values of $g$, because the right side goes to $-\infty$ as $d$ goes to $\infty$. (In particular, $g\leq 12$.)

Thus, one cannot embed all curves of large genus using only simple cusps. More complicated singularities could give larger reductions in the genus, at the price of comparatively larger reductions in the size of the moduli space. I don't think that can save you, but I don't know.

share|improve this answer
Dear Will, thank you for this answer. I agree that "one cusp on the curve" gives you codimension $2$. But I don't see why having $k$ cusps, when $k$ is around $(d+1)(d+2)/4$ gives you codimension $2k$. You need some transversality for this. Where would you get it from? –  aglearner Feb 22 '13 at 23:17
I can't think of an explanation. You certainly have transversality for $d$ sufficiently large with respect to $k$, since the map from functions to their values, first derivatives, and second derivatives at $k$ fixed points is surjective for large $d$. But I'm not sure how to make this effective or if that would be enough. However, I think it's a pretty good heuristic. You would need to get a lot of non-transversality to prove the reverse of this. –  Will Sawin Feb 23 '13 at 5:43
One more comment: there exist sextic curves of genus 1 with 9 simple cusps of multiplicity 2 (dual to a smooth cubic curves), but any rational curves of the plane has at most 8 cusps (Tono, K.: On the number of the cusps of cuspidal plane curves. Math. Nachr. 278 (2005), 216–221.) In particular, there is no rational sextic with simple cusps of multiplicity 2 (this is easy check in this case: look at the dual curve). This shows that imposing the conditions at cusps is really a global question. –  Jérémy Blanc Feb 23 '13 at 10:25
@Will: "The most typical type of singularity is a simple cusp" --- I would say that the most typical is a node. –  Sasha Feb 23 '13 at 18:52
For $2k \le d+1$ you have enough transversality to show codimension $2k$ (since $\mathcal{O}(d)$ is $(d+1)$-very-ample and therefore induces locally versal deformations of singularities with Tjurina number up to $d+1$), and always having $k$ cusps is at least codimension $k+1$, because the Severi varieties are of codimension $k$ and the nodal locus inside is open and dense. I don't know of better bounds. –  Vivek Shende Feb 23 '13 at 22:51

The question is probably hard to answer in general. Just an easy observation: let $g>0$ be an integer, take a curve of equation $y^2z^{2g}=p_{2g+2}(x,z)$, where $p_{2g+2}$ is a polynomial of degree $2g+2$ with only simple roots, and such that $z$ does not divides $p_{2g+2}(x,z)$. This is the equation of a curve of genus $g$ in $\mathbb{P}^2$, and the only singularity is the point $[0:1:0]$, which is a cusp. In consequence, you get an injective morphism from the resolution of the curve to $\mathbb{P}^2$. And all hyperelliptic curves are obtained by this way. At least, you see that there is no obstruction on the genus. But of course, a general curve of genus $g$ with $g>2$ is not hyperelliptic, so this is only a partial answer.

share|improve this answer
Jeremy, thank you! nice to hear that the question is not completely trivial –  aglearner Feb 23 '13 at 14:18

Your Answer

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.


text:
Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In a computer graphing library, a rectangular region of the Cartesian plane may be defined by {x, y, w, h} (where w,h are width and height). Intersection (lets say '^') is defined as the overlapping region of two rectangles (and also is a rectangle). Union of r1 and r2 could be defined as the smallest (ie smallest w, smallest h) rectangle 'u' such that u ^ r1 = r1 and u ^ r2 = r2

My question(s):

  • does negative w, h make 'sense',
  • if so , what sense does it make? ie, and in particular, should the result of intersection of non-overlapping rectangles be a rectangle with negative width and/or height?
  • can this be used to build a Group over the set of rectangles with the operation of union and/or intersection?

  • what is the 'area' of a rectangle with width < 0 and height < 0. I would like it to be negative, but typically it is width x height, thus positive.

Sorry if this sounds more computerish than mathish, but I was once a math major, and non of the other computerish people I might ask know anything about group theory...

More thoughts:

I want to decide if rectangles with negative dimensions makes much sense. I could leave this up to the users of the rectangles, but it is up to me to define what an operation like intersection does.

'neg' rectangles could just mean that the x,y is at the opposite corner than expected. ie they could be 'facing' the wrong way, and by moving x,y to the opposite corner, w,h can be made positive, and the rectangles are 'normalized'.

But when I consider intersection, imagine 2 rectangles moving along the plane, at first intersecting, but moving such that they overlap less and less - they eventually get to the point where they only touch at the corners - and the intersection is then a rectangle with w = h = 0.

Now, If we continue to move the rectangles further apart, you could either say that the result of intersection is still an empty rectangle, or I think it might make more sense to say the result is a 'negative' rectangle (and w,h depicts how far away they are from overlapping).

Given this concept, can it be somehow continued/followed/expanded in a logical/consistent way, and if so, what do we end up with?

Mathematically it would nice if it was a group, and if making it a group required a few other common operations (like union maybe) to be defined in a certain consistent way, I would like to explore that.

I just don't have any colleges nearby that think mathematically...

share|improve this question
Any definition that you could want to make depends on what you want the definition to do for you. What purpose do you want this to serve? –  Qiaochu Yuan Mar 1 '10 at 20:01
I'd say that what the union/intersection allows you to do is to define a lattice (i.e., a poset with finite meets and joins) in the set of rectangles. I haven't checked but you might get a monoid by considering the endomorphisms of the lattice, which you could then turn into a group using the Grothendieck construction. I don't see, though, how this is related to rectangles with negative dimensions. –  José Figueroa-O'Farrill Mar 1 '10 at 20:02
You are not going to get a group (independently of what you decide to do with negative widths and heights), because you already have problems with cancellation: the union of {x:0, y:0, w:1, h:1} with {1, 1, 1, 1} is the same as the union of {0,0,2,1} with {1, 1, 1, 1}, namely {0, 0, 2, 2}. A similar thing happens with intersections. –  Mariano Suárez-Alvarez Mar 1 '10 at 20:06
I'll update the quesiton with a few more thoughts... –  tony Mar 1 '10 at 20:11
I apologize if you were already aware of this - it is hard to tell from what you've said - but a group has just one operation. –  Zev Chonoles Mar 1 '10 at 20:18

1 Answer 1

up vote 1 down vote accepted

I imagine the only useful way of interpreting a rectangle with negative w and/or h should just be as a rectangle with positive width and height, only starting at a different point (it's lower left vertex, it seems?).

With either union or intersection as the operation, you're going to run into a couple of problems if you want the structure to be a group. The only possible choice for the identity under intersection is the entire plane (which, if you are parametrizing the rectangles in the plane using (x,y,w,h), would require you to allow x=y= $-\infty$, w,h=$\infty$), and the only possible choice for the identity under union is an "empty" rectangle (i.e., w=h=0, though you'll also have the problem that there is one of these at each x and y). Of course, given any rectangle which is not the entire plane, you can't take an intersection of it with something and get the entire plane, and similarly, given any rectangle which is non-empty, you can't take the union of it with something and get an empty rectangle - that is, there won't be inverses. However, as you guessed in the title, you still do get a commutative monoid.

You can make a group (and even ring) structure, similar to what you are describing, out of the collection of all subsets of the plane (or any set), as described here, but you'd have to give up your restriction to only rectangles if you want to use this one, because the operation used is the symmetric difference.

EDIT: This is to summarize the things you seem to be looking for:

  • The set of rectangles in the plane, under the operation of intersection, makes a commutative monoid if you allow for the "infinite rectangle" consisting of the entire plane.
  • Unfortunately, under union (as you defined it for rectangles - least rectangle containing both), there is in fact no identity, because given any two distinct "empty" rectangles, let's say the ones at (a,b) and (c,d), their "union" would be the rectangle with opposite corners at (a,b) and (c,d) and so is non-empty, so neither of them can be an identity. Thus, the set of rectangles in the plane, under the operation of "union", is only a commutative semigroup.
  • The set of rectangles is not closed under taking symmetric difference or the usual union.

That describes the situation with the main set-theoretic operations, though you may be able to construct another operation on rectangles using some combination of addition and multiplication on the parameters (x,y,w,h) which makes a stronger structure.

share|improve this answer
Interesting thoughts, thanks. –  tony Mar 1 '10 at 20:30

Your Answer

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.


text:
Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $(\Omega,P,\mathcal{F})$ be a probability space with filtration $\mathbb{F} = (\mathcal{F}_t), t \in [0,T]$, where $T$ can be finite or infinite. Let $M$ be a cadlag (local) martingale with respect to $\mathbb{F}$, and let $\mathbb{F}^M$ be the filtration generated by $M$ and then completed with respect to $P$.

Question: Is $\mathbb{F}^M$ a right-continuous filtration?

Some facts:

  1. If $X$ is a strong markov process, then the completion of $\mathbb{F}^X$ is right-continuous. This is in Karatzas and Shreve.
  2. A sort of converse: If $M$ is a local martingale in a right continuous and complete filtration, it has a right continuous modification.

One possible idea: A continuous local martingale can be expressed as a time-changed Brownian Motion, which is strong markov.

share|improve this question
It seems to me like an affirmative answer follows directly the "càd" part of càdlàg...perhaps I'm overlooking some nuance? –  Steve Huntsman Mar 3 '11 at 0:35
Youre saying the completed filtration generated by any cadlag process is right continuous? If things aren't completed, sets like {M has a local maximum at time t} are in $\mathcal{F}_{t+}$ but not in $\mathcal{F}_t$. –  weakstar Mar 3 '11 at 0:43
Ah, silly me... –  Steve Huntsman Mar 3 '11 at 1:05
For a brownian motion, those kinds of sets have probability zero. I'm not sure if that's the only kind of obstruction, but since martingales have the same paths as brownian motion, I had hoped you might be able to say something. –  weakstar Mar 3 '11 at 1:16

1 Answer 1

up vote 8 down vote accepted

No, that is not true. Consider the following, defined on a filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}\_{t\in[0,T]},\mathbb{P})$.

  1. $W$ is a standard Brownian motion.
  2. $U$ is an $\mathcal{F}_0$-measurable Bernoulli random variable independent of $W$, with $\mathbb{P}(U=0)=\mathbb{P}(U=1)=1/2$.

Then, set $M_t=UW_t$. This is a continuous martingale. If $\mathcal{F}^M_t$ is its completed natural filtration then $U$ is $\mathcal{F}^M_t$-measurable for all $t > 0$. Then, $U$ is $\mathcal{F}^M_{0+}$-measurable but is not measurable with respect to $\mathcal{F}^M_0$ (which only contains sets with probability 0 and 1). So $\mathcal{F}^M_{0+}\not=\mathcal{F}^M_0$.

Also, this is essentially the same as the example I gave in a previous answer of a Markov process which is not strong Markov.

As another example to show that there is not really any simple way you can modify the question to get an affirmative answer, consider the following; a Brownian motion $W$ and left-continuous, positive, and locally bounded adapted process $H$. Then, $M=H_0+\int H\\,dW$ is a local martingale. Also, $M$ has quadratic variation $[M]=\int H^2_t\\,dt$ which has left-derivative $H^2$ for all $t > 0$. So, $H_t$ is $\mathcal{F}^M_t$-measurable, as is $W_t=\int H^{-1}\\,dM$. In fact, $\mathbb{F}^M$ is the completed natural filtration generated by $W$ and $H$. If $H$ is taken to be independent of $W$, then $\mathbb{F}^M$ will only be right-continuous if $\mathbb{F}^H$ is, and it easy to pick left-continuous processes whose completed natural filtration fails to be right-continuous.

share|improve this answer
Thanks. Maybe i'm being dumb, but can this counterexample be extended to a nonzero starting position? –  weakstar Mar 3 '11 at 1:07
@weakstar: I'm not quite sure what you mean. Setting $M_t=UW_t+c$ extends it to a nonzero starting position. If $W$ was started from a nonzero position and you wrote $M=UW_t$, then $U$ would be $\mathcal{F}^M_0$-measurable, so the argument breaks down. –  George Lowther Mar 3 '11 at 1:26
Right, I meant your second suggestion, for brownian motion starting away from zero. Just seeing if any affirmative answer can be salvaged. –  weakstar Mar 3 '11 at 1:45
@ George Lowther : Hi, could you please recall the version of the Bluementhal 0-1 Law, you are using to justify that sets in $\mathcal{F}_{0^+}^M$ have 0-1 measure. Best Regards –  The Bridge Mar 3 '11 at 8:32
@The Bridge: I'm not using any such law, nor am I saying that $\mathcal{F}^M\_{0+}$ only has sets with probability 0 or 1 (it doesn't). I was saying that $\mathcal{F}^M\_{0}$ only has sets of measure 0 or 1 (in the first example) because it is generated by a random variable which is identically zero (and then completed). –  George Lowther Mar 3 '11 at 9:02

Your Answer

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.


text:
It’s a simple rule: When multiplying two numbers, if the signs are the same, the product is positive, if the signs are different, the product is negative.

But why does it work?

In the spirit of Khan Academy’s new and improved video, here is a lesson, in a context familiar to students, that builds upon the Distributive Property, multiplying by zero, sums of opposites, and substitution, in order to explain why the rule works.

Have students answer the following questions:

1) You are going to a new restaurant with your family. Tonight’s special is $7 for any dinner on the menu. Dessert is not included and costs $2 for each item ordered. If there are five people in your family, and each orders dinner and dessert, how much will the meal cost (ignore tax and tips)?

 2) The waiter brings you your bill. You pay it. What is your balance now for the meal?

 3) If you decide to eat at home instead tonight, what will your cost be for dinner from the restaurant?

Invite a student to share their strategy for question 1. Then ask if anyone solved the problem differently. Typically, some students will add to get the total cost of the dinner first and then multiply by 5 people, while others will first multiply to find the cost of 5 dinners and 5 desserts and then add. Discuss how this is an application of the Distributive Property. Use an array to show that, when multiplying two numbers, you can decompose one of the factors in any way and you still have the same number of total items no matter how you break them up. Practice this by challenging students to mentally calcuate, for example, 6 x 45, by decomposing 45 and writing 6 x (45) = 6 x (40 + 5) = (6 x 40) + (6 x 5) = 240 + 30 = 270.

Next use question 2 to review that opposites sum to zero. The situation can be described by writing – 45 + 45 = 0, and you owe zero dollars after you pay.

Use question 3 to discuss that multiplying by zero yields a product of zero. The meal costs $9 but zero people purchase it, so 9 x 0 = 0 cost for the restaurant meal. Also, since 9 x 0 and 0 are equal, we can just say 0 instead of 9 x 0.

Finally, use these concepts to show how the rules for multiplication can be generated:

We know that any number times zero equals zero.

So, for example, 5 x 0 = 0

We also know that opposites sum to zero. So we can replace the factor of zero with any sum of opposites:

5 x (-3 + 3) = 0

By the Distributive Property, this can be rewritten as

(5 x -3) + (5 x 3) = 0

Evaluating 5 x 3:

(5 x -3) + 15 = 0

And since we know that opposites sum to zero, then the addend (5 x -3) has to be equal to the opposite of 15, or -15.

Since we could have used any pair of opposites, our procedure is general, works for all numbers, and therefore the rule that a positive times a negative is negative must always hold true.

Next, try -5 x 0 = 0

Repeating the steps from above, first substitute a sum of opposites for zero:

-5 x (-3 + 3) = 0

Distributing, (-5 x -3) + (-5 x 3) = 0

We saw before that -5 x 3 = -15 so substitute

(-5 x -3) + -15 = 0

Since the addends sum to zero, -5 x -3 has to be the opposite of -15 or positive 15 and, since we can generalize with any pair of opposites, we have shown that the product of negatives must always be positive.


text:
Package org.apache.maven.plugin

Code to cache and manage plugin execution.

Interface Summary

Class Summary
GoalToJellyScriptHousingMapper This is the process that we use: - Start with goal X - Gather prereqs for X We want to find all goals that will be executed due to X being executed we don't care about ordering here because when werkz executes it will sort that out for us.
PluginCacheManager Plugin cache management.
PluginManager Plugin manager for MavenSession.
PluginScriptParser Class to parse plugin XML scripts to find goals and tag libraries.

Exception Summary
UnknownPluginException Indicates an attempt to access an unknown plugin.

Package org.apache.maven.plugin Description

Code to cache and manage plugin execution.

Copyright © 2001-2008 Apache Software Foundation. All Rights Reserved.


text:
Fork me on GitHub

Trouble Shooting

A list of more in-depth trouble shooting guides

The JDK being used is different than you expected.

After running mvn eclipse:eclipse your project's are using a different JDK than expected.

See jdk being used is different than expected.

The Eclipse project is not named correctly

Please see the FAQ Eclipse is not using the correct project name and is using the directory name instead.


text:
Click to go to Forum Home Click to go to maXbimmer Home

Go Back   maXbimmer Forums > Members List
User Name

Welcome to!

You are currently viewing our boards as a guest which gives you limited access to view most discussions and access our other features. By joining our free community you will have access to post topics, communicate privately with other members (PM), respond to polls, upload content and access many other special features. Registration is fast, simple and absolutely free so please, join our community today!

DINO K DINO K is offline

Sauber F1 Team

About Me

  • About DINO K
    My car
    E90 330i

Total Posts
General Information
  • Last Activity: 09-28-2010 07:37 PM
  • Join Date: 07-26-2009
  • Referrals: 0

Showing Friends 1 to 1 of 1

All times are GMT -4. The time now is 12:11 PM.

Powered by vBulletin® Version 3.8.1
Copyright ©2000 - 2015, Jelsoft Enterprises Ltd.
Maxbimmer Copyright 2001 - 2015


text:
Testimony of Harrison Livingstone

Hearing of 10/11/94 -- Washington, DC.
Our next scheduled witness is Mr. Harrison Livingstone.

MR. LIVINGSTONE: Thank you for inviting me. Also, I think it is terribly important that these hearings are open, and it is a very good thing that you have done this.

I a best selling author. I have written four books on the assassination of President Kennedy, two of them were best sellers, the fourth one will be published in the spring. I represent the Association of Assassination Researchers, which is new and just forming.

I think everything that people have discussed here today is quite important in various areas in terms of the records that they want to obtain, but I think you need priorities, and those priorities should start with certain facts in the evidence of the assassination.

In other words, where we are in our criminal investigation today, I do have the help of law enforcement on their own time. We do a great deal of many trips to Dallas, and very intense investigation. The focus of my personal investigation is in the medical evidence and the forgery of the autopsy photographs, the forgery of the X-rays. I am responsible for that, and we have basically broken the case. You need to take some time to study just where that is at because your priorities should include making every available document and piece of physical evidence connected with those forgeries, and the faking of the autopsy in President Kennedy's murder your primary responsibility.

As for your problem with defining just what is documents in this case, does that include the physical evidence, the Bullet 399, for instance, the rifle, the windshield as has been mentioned here. This will lead in a moment to the problems that I am having with the National Archives right now, an agency that you need to be sure that they are making documents and materials available, and they are not.

With 399, for very many years no one has been able to get access to it. Josiah Thompson described his efforts in Six Seconds in Dallas trying to obtain access with a set of very fine scales to weigh 399 and perform other examinations.

I am trying to get to see this material with trained police, forensic experts, police officers, to see this material, and I have been given a royal runaround, and have even been framed by staffers at the National Archives for saying and doing things that I could not have done because they don't want to show this material.

The reason why this is important to you is that if you look at the photographs of 399 published by the Warren Commission and have engineering drawings made of them, which I have submitted to the National Archives, that bullet is entirely incompatible with the alleged murder weapon, the rifle. The lands and grooves are not of the same count.

With regards to the autopsy photographs and X-rays, you need to examine the whole business of the agreement with the Kennedy family which has put this material out of your reach and out of our reach and where people have to go and get permission from Burke Marshall at Yale.

I understand the reasons at the time in 1963 for protecting the sensibilities of the Kennedy family. I am very aware of that. I worked for Senator Tydings who is very close to the Kennedy family. But at this point, 30 years later, people certainly have gotten over their grief, and we have to make this material available to it.

Your next priority is, you have to see what the forgery of that evidence means, and what the evidence is there for, and then all records connected with Bethesda Naval Hospital, with the development of the alleged autopsy photographs, with the X-rays, all military personnel records of every single person that might have had any connection with that autopsy or anything else, we need that material. We need to know where General Charles Cabell was on that day. We need to know everything there is that can be found out about Admiral Burkley, about Admiral Galloway, and all the rest.

Very briefly, before I get to two other problems here, if you, yourself, take the time to look at the interviews conducted by the House of Representatives with the autopsy doctors, they insisted that the autopsy photographs to not show the wounds on the body that they had, and the House of Representatives, as I pointed out in my first book, commented on the fact of the persistent disparity between the findings of the autopsy pathologist and what we see in the autopsy pictures. The wounds have moved quite a bit.

No, the body wasn't stolen; no, it wasn't altered; but, yes, all of the physical evidence in this case was faked, and this is the core of the mystery of what happened to John Kennedy.

Now we hear here today and with these organizations a lot of finger-pointing at everybody's favorite scapegoat, which is the CIA, and I am not here to protect the CIA, but the point is that all the attention is being directed away from the military, away from Bethesda Naval Hospital, away from the Office of Naval Intelligence, away from Army Intelligence, Military Intelligence, and this has been a common thing in this case for many years with attention being drawn to the Mafia, to Castro, to Russia -- Krushchev did it, everybody and his brother.

But the bottom line of it is that the medical evidence in this case is a fraud and a lie. If your priority should be, if you have an investigative capacity, to take a look at that and start dealing with it and make damned sure that every possible interview, every possible document that might be remotely with that coverup that began that night at Bethesda Naval Hospital that that material be made available.

The agreement with the Kennedy family is clearly an illegal agreement with regard to the autopsy pictures and X-rays and other materials because this is evidence in a criminal case and it cannot be sequestered and taken out of the public jurisdiction. It seems to me that you do have jurisdiction over that matter.

There was a suit here in Federal court which was dismissed to try to overturn that agreement. I hope that if you want, you will get in touch with me in the future on that matter.

Another issue that I want to briefly talk about is, what is going on at the National Archives, an agency. Presumably you are dealing with "agencies," so far we hear about the CIA and the FBI and so on, but the National Archives is an agency and there is a kind of surveillance going on now that these documents are being made available that is unconscionable. It is way over the line between reasonable need to protect these materials and deliberate harassment of researchers.

MR. LIVINGSTONE: Thank you. Anna Marie stormed out of here a little while ago, she was driven to the point of hysteria in the National Archives, and staffers are attaching themselves to researchers and will not go away. The engage you in long conversations, and I clocked one of these conversations for five hours, and researchers flew from California and a staffer devoted himself to distracting that researcher from why he had spent all the money to come here. So he didn't get what he was after. He was engaged in a social discussion, a political discussion, and so on. The staffers are finding out what people are researching. It seems to be floating around and being reported to other people.

Before going any further, I want to say that this does not mean Steve Tilley because he has been nothing but nice to me and everybody that I know, but we also know of other cases there where the exact opposite is happening, and I have personal knowledge of people actually being framed for things in order to get rid of them. Anna Marie was threatened with the loss of her research privileges, and so on.

So it is a conflict between reasonable surveillance for the protection of the documents in the National Archives and where does it cross the line into harassment.

Now you need to define, is this physical material, the bullet, the rifle, the windshield, and all that, is this -- I don't know -- I don't think it is specifically defined in the Act, so in my view you need to define that as part of "documents," or whatever.

My final quick point is just, the National Archives should be encouraged to make slides of each and every frame of all the other films that might be available, such as the Nix film, the Hughes film, the Bronson film especially, and to do blow-ups of this, and the Secret Service copies of the Zapruder film which is badly damaged, you should inquire as to why it is so badly damaged with the Secret Service. It is broken and ruined in a lot of places. How could this have happened? How could they take such critical evidence in this tragedy and it be ruined? But this visual evidence in this case is the key to the murder of John Kennedy because it is all fake, and I mean the Zapruder film is a fake.

The autopsy photographs are a fake. The autopsy X-rays are a fake. We first determined this in 1979. In September Dr. David Mantik proved it with a test in physics. My best friend is a chief radiologist, we have been examining this for many years. We went to college together, and he is going to take that study to the next step. The photographs have been wildly denounced by the autopsy doctors themselves, but those documents are clearly edited. The documents that were released here last fall are clearly edited.

At some other time, I will be glad to discuss that with you. I won't take any more of your time.

CHAIRMAN TUNHEIM: Thank you, Dr. Livingstone.

CHAIRMAN TUNHEIM: We appreciate you coming today.

Search Go to testimony Go to Fair Play Releases E-mail ARRB Go to home page ARRB Info Top of ARRB page