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text: The Chicago resident Joe Swanberg and the New York-based Greta Gerwig directed and star in this "mumblecore" romantic drama as James, a video-game designer who lives in Chicago, and Maddie, a New York nursing student who writes in her spare time. In the opening long take, as they enter James's apartment, they strip each other in a minute and nineteen seconds, giving rise to a performance by Swanberg that is hard to fake. The humid intimacy makes clear what the artists think love is based on, but it soon becomes clear that the couple's long-distance bond is strained by their absence from the wider web of each other's life. As portrayed by the inventive Gerwig, Maddie, who is whimsically articulate and quietly tempestuous, is in command of the relationship; Swanberg makes James merely a man—stolid, ambitious, capable, sentimental, and self-absorbed. The seemingly improvised dialogue and unblocked staging lends emotional weight to their fleeting glances and sudden gestures, and such moments as a brief, inspired, tender streetside dance blossom exquisitely in real time. Gerwig and Swanberg rely heavily on their cameraman, who sometimes gets close enough to celebrate their beautiful imperfections but usually keeps the onscreen couple at a poignant distance and in ambiguous equilibrium; their raw, highly personalized methods convey the sense that the lives depicted on-camera and those lived behind it are one.


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text: I've searched everywhere for someone with a similar problem to what I have, but no one seems to have the same problem. So the problem I'm having is that my MacBook pro won't charge and it won't run off the wall adapter, the MagSafe has a very dim green light on it, and the computer won't recognize the fact that the charger is even there, but on occasion the Mac will begin to charge and the MagSafe will become amber but if I disconnect the charger when it's charging and then reconnect it it stops charging. (I've rest the pram and this seems to have made the charger green but the computer still doesn't recognize it.) Now that you know what is physically wrong let me tell you how it got this way in the first place. Back in March I spit Mountain Dew on my Mac, since then I've replaced the logic board and DC board after that the Mac had no problems. Until OS X Mavericks develop release came out, I couldn't help myself so I installed it, it was all fun and games until the Mac have the problem I just described, I fixed it by restoring from a time machine backup that put me back to OS X Mountain Lion, it continued to work for 3 1/2 weeks until the problem arouse again. I "fixed" it this time my going to the restore page (Command-R) but yesterday the problem came back, and I've restored to Lion and Mountain lion, I've done a clean install, and still nothing. I'd like to thaw it to a service provider buttons closest one is 3 hours away and I could probably fix it myself seeing how I've fixed my Mac on multiple occasions before this, all I'm wondering is: what could be the problem?


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Date: January 31, 2011
On November 8, 2010, all states and territories received notification from the U. S. Department of Education's Office of Special Education Programs (OSEP), that all states making application for funds under the Individuals with Disabilities Education Act (IDEA) are required to make available for public review and comment the application packet for the federal IDEA Part C funds. The Department of Elementary and Secondary Education has developed the draft application packet for these funds and is by this notice making the application available for review and comment at: http://dese.mo.gov/divspeced/Compliance/Part-C/index2.html.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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#1 2006-07-13 10:06:19
For a time I wondered how one could possibly obtain exact values for various zeta functions, such as ζ(2) = π2/6 and ζ(4) = π4/90. Then, one fateful day, by luck or extreme insight, I realized that Fourier series could possibly do the job. Upon testing, I was pleasantly led the to correct answer. Due to the nature of the method, only values for ζ(2n) are obtainable. Of course, as far as I know, ζ(2n + 1) (odd values, basically) always appears unable to be expressed in terms of known constants. Anyway, let us move on to the method.
For this problem, we will have f(x) = x2 and L = 2π.
The the Fourier series expansion for f(x) is
Now when x = 0, the series becomes
Now, for the convergence of Fourier series, there are 3 conditions called the Dirichlet conditions:
1. f(x) is defined, expect for possibly at a finite number of points in (0, 2L).
2. f(x) is 2L-periodic outside (0, 2L).
3, f(x) and f'(x) are piecewise continuous in (0, 2L).
Now, our function is definitely defined all throughout (0, 2L). We defined f(x) as being 2L periodic, so it is a parabola from x = 0 to x = 2π, and this is repeated periodically(so f(x + 2π) = f(x)). On (0, 2L), f(x) is a continuous parabola and f'(x) is a continuous straight line. So the Dirichlet conditions are satisfied.
Now that the Dirichlet conditions are known to be satisfied, what does that do for us? It does something very good. It tells us that the Fourier series for f(x) will converge to
Now let us return to the problem at hand. We had just let x = 0. This means we can find what the Fourier series will converge to when x = 0, thanks to the Dirichlet conditions:
Now we may write the Fourier series at x = 0 as
Now we may easily solve the equation for ζ(2):
And so it is done. In general, to find ζ(2n), write the Fourier series for a 2π-periodic function f(x) = x2n defined on (0, 2π) and work out the problem as I have above.
Does anyone else have methods for finding exact values of infinite series?
Last edited by Zhylliolom (2006-07-13 10:10:36)
#2 2006-07-13 10:51:33
Re: ζ(2n).
Sure, there are many ways. Alternating series can be pretty easy. For example:
I already gave you a major hint that it is an alternating series.
But in general, what you try to do is take the first few summations, and come up with a sequence. Then you must prove that it actually is the sequence through induction (which normally isn't too bad). Then you just have to show what number that sequence converges to, which can be very painful or even just about impossible.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#3 2006-07-13 15:04:39
Re: ζ(2n).
Ricky, your sum is divergent. Is i supposed to start at 1 instead? I'll assume this is the case and try out the problem. Also, I don't know what you mean by an alternating series, don't alternating series have (-1)i in them?
After these first few partial sums, a pattern is noted:
We will now prove this by induction. We start with the base case:
So the base case holds. Now assume that
Then we now want to show that the equality holds for n + 1:
So by mathematical induction,
Now the question is, "what do the partial sums converge to?" To find this out we take the limit as n approaches infinity:
So we conclude that the series
converges to the value of 1.
Last edited by Zhylliolom (2006-08-09 19:58:03)
#4 2006-07-13 23:35:41
Re: ζ(2n).
I didn't really read through your proofs, but you got the answer right. But there is an easier way to do it. As I said, this is an alternating series, though it doesn't look like it.
Now when you write the sum, it will look like:
But we can't reorder terms until we prove that the series is convergent. So let's prove the series is convergent:
And now we're done. We can regroup terms, and so the sum is 1, since all the fractions cancle themselves out.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#5 2006-07-13 23:55:06
Re: ζ(2n).
Great I can understand #3, and #4 both.
#6 2006-07-14 00:04:37
Re: ζ(2n).
Fourier series are so difficult.
#7 2006-07-14 08:04:12
Re: ζ(2n).
Ricky: I feel a little silly now that I didn't approach it that way. But I arrived at the same answer, so I suppose it isn't too bad. I like your method better though.
#8 2006-07-14 09:15:58
Re: ζ(2n).
You need a good book on real analysis. In my experience, infinite sums are either really easy, or almost impossibly hard.
Great! Now you know how I felt when I had my advanced calculus final a while back with this exact problem on it, and I was the last one in the room with only this problem to go.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#9 2006-07-25 22:50:01
Re: ζ(2n).
The discovery of another method of finding ζ(2) has led me to post on this topic again. I find this one enjoyable, as it doesn't seem like a standard approach.
Now divide by x to obtain
This function will have zeros at nπ, so factor it according to that fact:
This simplifies to
Now if some crazy old man were to multiply this all out, the coefficient of x² would be
But in our original expression for sin(x)/x, the coefficient for x² was -1/3! = -1/6, so the sum must be equal to -1/6:
Now, simplify:
#10 2006-07-26 01:56:01
Re: ζ(2n).
It most certainly isn't standard. How in the world did you come up with that?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#11 2006-07-26 16:12:05
Re: ζ(2n).
I know nothing about zeta functions
#12 2006-08-09 19:56:41
Re: ζ(2n).
Great news, folks. After a haunting dream and a good portion of the evening spent sitting furiously with my notebook and pencil, I have found a simple equation which can be used to calculate ζ(2n). No need for Fourier series or wacky factoring approaches, this formula works for all positive integers n. After finding this equation, an immediate consequence was noted: it could be used to find ζ(-n) as well, and also I will show how this miraculous formula can be used to prove a fundamental property of the zeros of the Riemann zeta function(no skipping ahead to see this part, let the suspense build).
There isn't really a simpler way to define them, aside from a definition using a contour integral. The first few Bernoulli numbers are
From those few values you can probably notice one thing: for integers n > 0, B2n + 1 = 0, or in other words, odd values of n greater than 1 for Bn give the value zero. This concept will be important much later in this exposition, so try to remember it. Of course, you will be reminded again anyway.
isn't really in a form we can play with much for our present situation. Let's convert it into something we know more about. We can add x/2 to it and get something decent:
So to summarize,
Now we try to work something out for coth using our knowledge of Bn's generating function:
Now we already established that for all odd positive integers n > 1, Bn = 0. Thus we can make the substitution n = 2k in the above sum given that we make up for the loss of the term corresponding to n = 1. A quick calculation shows that the term we need is equal to -x/2, so we must add this in for our substitution n = 2k to work:
Alright, great. Now we would prefer something more familiar than the hyperbolic cotangent to work with. Let's use the substitution x = 2iz to get our standard circular cotangent(note that coth ix = -i cot x):
Beautiful series there. But this is no time for sightseeing, we have an equation to derive. Let us consider the known identity(whose proof is left to the reader as an exercise )
Let's clean up the sum a little to make it more workable. Start by making the substitution x = z/π and multiplying through by it:
Now let's push our manipulation of this even farther:
An observant reader may notice that our summand is actually the value of the geometric series
Because of this fact, we will substitute this sum in for the summand and upon manipulation make a marvelous discovery:
So, after all that work, we now have two sums for z cot z, and we may write
Let's kick z cot z out of the picture now, as we have used him as much as we need to:
Now we want to get rid of the summations. It turns out that such a task is easy. We want the limits of summation to be equal, so let's get the lower limit on the left sum to be k = 1. We can do this by simply writing the lower limit as k = 1 and adding the value corresponding to k = 0 to the left side. The corresponding value is 1, so we now write
Now just subtract 1 from each side:
The summands must be equal now:
Solve for ζ(2k) to get
I prefer the expression with the imaginary unit, but some may be more comfortable rewriting it as
So there it is, a nice little formula for ζ(2n). But that was only part of what will be discussed in this post. Let's take things a step farther and find a formula for ζ(-n) as well! To do this we'll use our newly discovered formula for ζ(2n) and the functional equation for the zeta function,
Make the substitution s = 2n:
Use our formula for ζ(2n) and simplify:
Letting k = 2n we get the form
Now substituting n = k - 1, we get
Now we have the power to evaluate the zeta function for negative integers! Here's a few values:
We can also plug in n = 0 to get
Now, as promised, I will prove a fundamental property of the zeros of the Riemann zeta function. The Riemann hypothesis states that for s ≠ -2, -4, -6... such that ζ(s) = 0, Re(s) = 1/2. Let's use our nice equations to figure out why the hypothesis is stated the way it is, and to prove something about ζ(s) = 0.
The statement of the Riemann hypothesis disallows values of s that are negative even numbers. Why is this? Using the derived equation for ζ(-n), we can see that for negative even integers, the Riemann zeta function always seems to have a zero. These zeros must be kept out of the hypothesis because they obviously have real part not equal to 1/2. But one may ask this question about ζ(-2n): "Is it always zero?" The answer is yes, and I shall prove it for you.
From our formula for ζ(-n), we can write ζ(-2n) as
A long while ago, I told you that for all integers n > 0, B2n + 1 = 0. So if that is true, ζ(-2n)'s denominator will always be zero and thus ζ(-2n) will always be zero. So our task now is to prove that B2n + 1 = 0 for all integers n > 0.
Recall the expression
from earlier. We know that B1 is an odd value of n in Bn which does not equal zero, so we take its term out of the series and write a little note by our sigma:
Now give x a negative value in the equation and note that coth(-x) = -coth x:
It turns out that x coth x is an even function, and since it is equivalent to the sum, the sum must also be an even function. Thus we require that (-1)nBn = Bn, which only holds for even n, or in another way of stating the situation, all odd powers of x in the sum must have a coefficient of 0. Thus all B2n + 1 = 0 for integers n > 0 and ζ(-2n) is always zero.
Last edited by Zhylliolom (2006-08-13 08:10:31)
#13 2006-08-09 20:04:23
Re: ζ(2n).
reading that makes me dizzy sorry
Presenting the Prinny dance.
Take this dood! Huh doood!!! HUH DOOOOD!?!? DOOD HUH!!!!!! DOOOOOOOOOOOOOOOOOOOOOOOOOD!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
#14 2006-08-10 00:46:53
Re: ζ(2n).
Just need a bit of help with your variables. k, n are natural, x is real, and z is complex, right?
Is the identity supposed to be:
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#15 2006-08-10 00:56:11
Re: ζ(2n).
IPBLE: Increasing Performance By Lowering Expectations.
#16 2006-08-10 08:38:54
Re: ζ(2n).
Ok, I admit I rushed the post a little(I was excited to share this with everyone), so a few things might be confusing. In my notes, I used z and s (Riemann style) for the variables. Consider x and z to be both complex. Should I change x to s in my original post?
No, I did a few steps in one there(probably the only part of my proof where I made a small jump, I think). I substituted x = z/π to get this:
Then I multiplied through by x = z/π to get
Next, z²/π² was moved inside the sum, and inside the sum we clear fractions by multiplying by π²/π²:
So I think the part that confused you there was me getting from π cot πx to z cot z, but as you can see, the substitution cleared the π and gave us a z inside the cot, and the multiplication of both sides by z/π put a z outside the cot and cancelled out the π outside, giving us z cot z.
Last edited by Zhylliolom (2006-08-10 10:24:59)
#17 2006-08-10 18:16:29
Re: ζ(2n).
i give up i cant understadnd anything ur saying
Presenting the Prinny dance.
Take this dood! Huh doood!!! HUH DOOOOD!?!? DOOD HUH!!!!!! DOOOOOOOOOOOOOOOOOOOOOOOOOD!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
#18 2006-08-10 21:47:59
Re: ζ(2n).
Zhylliolom, or maybe someone else could explain what happened here, I'm a bit puzzled :
I noticed the geometric sum after that, but I've never worked with double sums before, ever. But does it basically mean that you take the sum over the sum, as in (k=1 to ∞) first and then (n=1 to ∞) over that previous sum?
If so, I learned something new by reading your post
Bang postponed. Not big enough. Reboot.
#19 2006-08-11 00:33:43
Re: ζ(2n).
numen, he took a negative sign out of the inside of the sum, and put it on the outside.
And as for the 2nd part, in short, yes.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#20 2007-05-12 01:30:07
Re: ζ(2n).
Zhylliolom wrote:
I give up, would you give me a hint?
#21 2007-05-12 01:31:27
Re: ζ(2n).
Zhylliolom wrote:
I give up, would you give me a hint?


text:
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In the News
Original journalism from the Center's OpenSecrets Blog
- Jack Abramoff's Money Still Casts Shadows as Fallen Lobbyist Is Transferred from Federal Prison to Halfway HouseJune 9, 2010 5:01 PM
Disgraced former lobbyist Jack Abramoff is out of the slammer, CNN has reported. Four years ago, Abramoff pled guilty to several charges related to defrauding millions from Indian tribes he represented as a lobbyist and corrupting public officials.
- Money and Votes Aligned in Congress's Last Debate Over Bank RegulationSeptember 23, 2008 11:43 AM
The last time Congress seriously debated how to regulate the financial industry, the result was legislation that allowed the nation's largest banks to get even larger and take risks that had been prohibited since the Great Depression. A look back at that debate, which was over the 1999 Financial Services Modernization Act, reveals that campaign contributions may have influenced the votes of politicians who, a decade later, are now grappling with the implosion of the giant banks they helped to foster.
- Abramoff: More Time Behind Bars and Former Colleague Ring IndictedSeptember 9, 2008 4:19 PM
The criminal investigation surrounding disgraced lobbyist Jack Abramoff continues to unfold. Last week--during the Republicans' convention--Abramoff, a staunch Republican who gave more than $200,000 to the GOP, was sentenced to four more years in prison, and on Monday Kevin Ring, who lobbied with Abramoff, was indicted on public corruption charges.
- Retirement FundsFebruary 21, 2008 4:22 PM
For congressmen who have been in office for decades, fundraising is easy and campaigning is cheap. And when they decide to leave office, donors' contributions can finance an extended retirement party.
- Rep. John Doolittle, Associate of Jack Abramoff, Has Retirement PlansJanuary 10, 2008 2:56 PM
A congressman under investigation for his ties to convicted lobbyist Jack Abramoff is expected to announce today that he will retire from the House of Representatives at the end of his term.
Feel free to distribute or cite this material, but please credit the Center for Responsive Politics. For permission to reprint for commercial uses, such as textbooks, contact the Center.


text:
Several hours before dawn on June 6, 1993, two Park Service police officers were patrolling the road next to Jacob Riis Park, a long stretch of beach on the Rockaway peninsula, in Queens, when they were startled by two Asian men flagging them down. As the officers got out of their car, they heard the sound of screams coming from the beach. The moon was full, and about a hundred yards offshore the officers saw a hundred-and-fifty-foot tramp steamer that had run aground. The ship's deck was crowded with people, and, as the officers watched, men and women jumped over the side, falling twenty feet into the surging waves below. Dozens of figures bobbed in the water, some managing to clamber ashore, others flailing wildly, apparently unable to swim. The officers radioed for backup.
The ship's name, stencilled in white block letters on the bow, was the Golden Venture. Its cargo was nearly three hundred illegal Chinese emigrants. Before reaching the Rockaways, the ship had sailed some seventeen thousand miles, from Thailand to Kenya, around the Cape of Good Hope, then across the Atlantic to New York.
The passengers—mostly adults, but a few children—were emaciated. They had been confined in the ship's hold for months, subsisting on rice, peanuts, and purified salt water. It had been uncomfortably hot, and many passengers wore only underwear; when they hit the water, which was fifty-three degrees, some went into cardiac arrest. One Coast Guard officer who performed CPR on two men onshore recalled, "I could feel the gristle of their bodies, the cartilage. They walked up out of the water, collapsed on the beach, and died."
Six bodies were recovered from the surf; four others were found later. By dawn, news helicopters were capturing live footage of the disaster. The Golden Venture accident was not an isolated incident: in the preceding year, more than a dozen ships had dropped human cargo from China on American shores. In April, a ship called the Mermaid 1, carrying two hundred and thirty-seven illegal Chinese, had been intercepted by the Coast Guard near the Bahamas. In May, the Pai Sheng had slipped beneath the Golden Gate Bridge at night, depositing two hundred and fifty passengers on a San Francisco pier. An internal Department of Justice report declared an "immigration emergency"; the San Francisco Chronicle heralded a "SMUGGLER SHIP INVASION."
Several miles from the beach, in a small shop at 47 East Broadway, in Manhattan's Chinatown, a middle-aged woman named Cheng Chui Ping watched the story unfold on television. Short and stout, with cropped black hair, wide-set dark eyes, and a hangdog expression, she was known in the neighborhood as Ping Jia—Sister Ping. Her gruff demeanor and simple clothes gave her the appearance of a Chinese peasant; she had little formal education, spoke almost no English, and spent most of her waking hours managing the shop, which sold clothes and goods from China, and a restaurant in the basement. But Sister Ping was also an extraordinarily wealthy businesswoman who owned the restaurant and the shop, as well as the building that housed them. She was what the Chinese call a shetou, or "snakehead"—an underworld entrepreneur who charges tens of thousands of dollars to shepherd undocumented migrants from one country to another. She helped purchase the Golden Venture, and two of its passengers owed her fees. One of them had died.
Last summer, twelve years after the Golden Venture ran aground, Sister Ping—now fifty-six, and a grandmother—was tried in a federal courtroom in New York City. She became the twenty-third person to be convicted in connection with the voyage. Described by the authorities as "the mother of all snakeheads," she was charged with operating what prosecutors called "a conglomerate built upon misery and greed." Moving people illegally from one country to another requires an extensive network of international contacts and an ability to outwit immigration and law-enforcement officers. With a well-connected family, acute entrepreneurial instincts, and a callous, life-is-cheap attitude toward the poor migrants who were her customers, Sister Ping was well suited to the job. Working with associates in China, Hong Kong, Thailand, Belize, Kenya, South Africa, Guatemala, Mexico, and Canada, she helped create the China-to-Chinatown route in the early nineteen-eighties and ushered thousands of undocumented Chinese emigrants to America. According to the F.B.I., over the course of two decades she made some forty million dollars.
Not long ago, I visited a man named Michael Chen, who arrived in America on the Golden Venture and had agreed to talk with me about the journey. Chen is now thirty-two. He is short and slight, with a boyish face and thick, expressive eyebrows. We met in the spotless Chinese restaurant that he owns in a strip mall in suburban Columbus, Ohio; in the lull between lunch and dinner, he related his ordeal.
The first Chinese who came to America, in the mid-nineteenth century, originated from a few counties on the Pearl River delta, around the southern city of Guangzhou. Michael Chen was part of another great wave of emigrants, who came from a series of villages along the Min River in Fujian Province, a mountainous sliver of coast across the straits from Taiwan.
Chen grew up outside Fuzhou, the regional capital. His father was a produce farmer who fished at night to supplement his income. Several of his uncles had gained entry to New York in the nineteen-eighties by paying snakeheads. These smugglers had emerged during the nineteen-sixties and seventies, when many mainland Chinese were fleeing to Hong Kong, which was then still in British hands. At Sister Ping's trial, one Fujianese snakehead explained that when smuggled emigrants slither through the wire fences strung along borders "the shape of it looks like a snake."


text:
April 24, 2011
Popeater recently spoke with former Beverly Hills 90210 star Jason Priestley and asked him if he had any advice for Robert Pattinson in regards to dealing with the whirlwind fame he's achieved:
What advice do you have for Rob Pattinson, who is the current idol of the moment?
It's cyclical. There's always a next one coming. I don't know if there's any advice, you just have to weather it, weather the storm and know that this too shall pass.
Via Diario Twilight


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If waterparks are a little too fomal or busy for your vacation style, here are some natural water holes that will soothe your travel-weary feet and soul.
Blue Ridge Highlands | Coastal Virginia - Eastern ShoreChesapeake Bay | Northern Virginia | Shenandoah ValleySouthern Virginia | Heart of AppalachiaCoastal Virginia - Hampton Roads
Cascades Recreation Area in Jefferson National Forest includes a two-mile trail with views of the 66' waterfall. The 14' foot pool at the bottom is inviting to swimmers, but it's icy cold year-round! Dip your toes or jump right in, if you dare.
Fairy Stone State Park features a 168-acre lake adjoining Philpott Reservoir just minutes from the Blue Ridge Parkway. This park is famous for its fairystones, which are little cross-shaped crystalline stones.
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Lake Rawlings - a SCUBA & Camping Park features very clear lake water for scuba diving, swimming and snorkeling. Who knows what you'll find on your underwater adventure!
You'll find Cave Springs, a small spring-fed swimming lake, in Wise County. The water is clean, crisp and definitely cool, as the lake is formed courtesy of a cave spring in the mountains above.
Another cold water lake is the 4-acre High Knob Lake with its 300' sandy swimming beach! Picnic, enjoy the bath house with showers, and keep an eye out for some entertainment at the nearby amphitheater. Camping available, too.
Buckroe Beach borders the Chesapeake Bay in Hampton and offers eight acres of family-friendly sandy beach. Rent chairs, umbrellas, and paddle boats for a full day of fun. Food vendors are often on-site with hot dogs and sno cones.
Huntington Park is a 60-acre park with a public beach on the James River in Newport News. Lifeguards are on duty and concessions are available daily during the summer.
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We are delighted to announce the club has been nominated as a finalist for the 2013 Starcount Social Star Award in recognition of the club's use of social media.
Starcount is a specialist website which ranks the world's top brands based on their popularity and fan engagement, to produce the ultimate social media chart.
With a social media following of nearly 25 million fans across the globe, Chelsea has been identified in the Sports category alongside Barcelona, Real Madrid and Manchester United as being the very best at engaging fans across social channels.
The club will find out if we have won the top award when all Social Star global category winners are unveiled on May 23 in Singapore.
The official Chelsea website will of course bring you the very latest, and you can learn more about Starcount, learn how to become a Superfan and view the other Social Star award nominees here.


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Physics Department Award Winners
October 2010 - Dr. Joshi and Dr. Daniels received a CFR Grant to start a project on Optical Tweezers.
September 2010 - The Department of Physics was selected as the recipient from the College of Sciences for the "Excellence in the Use of Technology" Award presented annually by the Center for Academic Technology Support (CATS).
July 2010 -
e." Dr. Zou also receives a grant to help with this project.
March 2010 - Dr. Jie Zou received a Council of Faculty Research Summer Grant.
March 2010 - Dr. Jim Conwell was awarded the 2010 COS Student Advisory Board Outstanding Faculty Award. This award is selected by students and honors Dr. Conwell's commitment and quality of teaching.
March 2010 -
The Physics Department was awarded the Provost's Award for Assessment. The award was a second place for outstanding achievement in program assessment.
February 2010 -
January 2010 - Robert E. Holmes, Jr., adjunct EIU Physics faculty, was the recipient of the 2009 Edgar Wilson Award for the discovery of comets by amateurs during the calendar year ending June 11, 2009. He discovered comet C/2008 N1 on 2008 July 1.
December 2009 - Dr. Zou and Dr. Joshi each won a College of Sciences Early Research Support Grant to aid in their research efforts.


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Free Mounting & Balancing: Professional mounting & balancing is available on all wheels and tires when purchased together on the same order, so you can just swap your wheels in your driveway when they arrive. We even offer Free Nitrogen tire fill (see option above).
- FREE Mounting and Balancing in 48 hours
- State of the Art Onsite Tire Facility
- Scratch-Free process + Hidden Weights
- TPMS Sensors installed free on 07+ Mustangs (see wheel page for details)
- Free Shipping on any order to the continental 48 states
- Please allow an extra 1 to 2 business days on your wheel and tire package for mounting and balancing
What is TPMS? The Tire pressure Monitoring System is a safety feature on all 2007+ Ford Mustangs that alerts you when your tire pressure drops to an unsafe limit.
How do I keep my TPMS feature? Aftermarket wheels don't have TPMS sensors. If you select 'Add TPMS Sensor Kit' and 'Add TPMS Tool' at the top of the page when you order your wheels, then chose Free Mounting & Balancing when you add tires, we'll install the sensors for free!
- Save Money - we sell at a fraction of Ford's price and install for free
- Save Safe - keep your safety feature intact
- No warning light - lose the annoying dash light that appears if your sensors are missing
For more Information, check out TPMS Sensor Kits or TPMS Tool in detail.
Mustang Fitment: 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014 Details >>


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Updated Dec 12, 2012 - 9:47 am
Report: Guard Mike Gibson to join Cardinals
Center Lyle Sendlein tore the MCL in his left knee during the team's Week 12 loss to the St. Louis Rams.
With Sendlein -- who had previously started 80 straight games -- forced to the sidelines for the remainder of the season, Ken Whisenhunt has relied on third-year veteran Rich Ohrnberger to start at center during the team's past two contests.
Ohrnberger, who had played in all 13 games this season, injured his knee in the Cardinals' 58-0 loss to the Seattle Seahawks Sunday and was placed on injured reserve Monday. Arizona replaced his roster spot with veteran quarterback Brian Hoyer, but still wanted some depth over the final three games. And from the looks of it, they may have found it Tuesday.
News: Cardinals signed OG-C Mike Gibson to help their horribly leaky offensive line today.— Eric Edholm (@Eric_Edholm) December 11, 2012
Gibson, a former sixth-round pick by the Philadelphia Eagles, has played in just 17 career NFL games. The guard last appeared with the Seattle Seahawks during the 2010 season.
Gibson was previously cut by the Eagles in training camp.
I'm back� glad to be apart of a team again. I am happy to be apart of the AZ cardinals organization.. #GoCards— Mike Gibson (@gibysmalls) December 12, 2012
- The Pulse: Cardinals fans most excited to see division rivals
- A disappointed Darnell Dockett: Karlos Dansby 'chased the money'
- Arizona Cardinals 'bought right in' to offseason strength and conditioning program
- Arizona Cardinals' Dockett: 'Anything less than the playoffs is a failure'
- Thoughts on the 2014 Arizona Cardinals schedule
- Michael Bidwill, Cardinals president - Thursday April 24What goes into the Cardinals� draft preparation?
- Mark Schlereth, ESPN NFL analyst - Tuesday April 22ESPN NFL analyst Mark Schlereth talks about the NFL Draft.
- Terry McDonough, Cardinals VP of player personnel - Tuesday April 22Who will the Cardinals draft with their first-round pick?
- Larry Fitzgerald, Cardinals wide receiver - Tuesday April 22How are the Cardinals getting better in the offseason?
- Derek Carr- Fresno State quarterback - Monday April 21Fresno State quarterback Derek Carr talks about the NFL Draft.


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Low Flat-Rate Shipping; Free on Orders Over $75 / 365-Day Returns and Free Return Shipping on All Skates
|Bauer Team Premium Hockey Bags 30"
Lowest Price Guarantee
The Bauer Team Premium Hockey Bag is your typical team bag, with an upgraded waterproof polyvinyl exterior for those wet locker rooms.
This bag has a large main compartment with heavy-duty double zippers with nylon pulls on the main opening. Bauer gave most bags in the 2012 line up a slightly taller and shorter shape, which gives the bag maximum "packability". Inside the main compartment, the bag has an inner accessory pocket great for tape, wax, stones, tools etc.
The bag has durable reinforced double layer webbed grab handles on the ends of the bag for easy loading/ unloading, and longer straps on the top of the bag for an easy cary to and from the rink.
- Model Number: 1040735
- Measurement: 30" L x 18" H x 15" W
- Material:
- Durable waterproof polyvinyl
- Graphics:
- Silk screened Bauer logos
- Main Compartment:
- Large main compartment
- Heavy-duty double zippers with nylon pulls on main opening
- Small accessories pocket located inside of the bag great for tape, wax, stones tools etc.
- Ventilation grommets
- Pockets:
- Internal accessory pocket
- External name/player number window
- Handles and Straps:
- Durable reinforced double layer webbing grab handles and straps
- Longer straps for easy carry


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Mother's Day Pop-Up Card
Age Range: 6-8
Craft Time: Under 30 mins
Messy Rating:
2 3 4 5
Uses Recycled Items: No
Get exclusive offers about this product and more!
Material List
- Printable Pattern
- Cardstock
- Crayola® Markers
- Creatology™ Chenille Stem - Green
- Hirschberg Acrylic Jewels
- Scissors
- Craft Smart® All Purpose Glue
Developmental Skills
- Fine Motor
- Visual/Spatial
- Virtue
Project Instructions
- Print pattern A onto white cardstock and pattern B onto red cardstock. Color designs as desired and cut out shapes.
- Fold heart shape from pattern A in half, leaving the design on the outside and cut the bold diagonal line.
- Fold card the opposite direction and open card half way. Use your finders to push the kite shaped piece forward to create the pop-up vase.
- Glue red heart from pattern B to the back of the printed heart as shown in photo. Let dry.
- Cut two 1½ inch pieces of green chenille. Glue a flower to one end and the other to the inside center of vase. Bend chenille pieces.
- Adhere jewels where desired with glue.
TipsCreate this cool pop-up card just for Mom!
To insure proper folding, only glue two flowers inside and make sure all glue is dried.