Recent content by camcool21

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    Why Did MIT Sever Ties with Walter Lewin?

    I think it is pretty clear that MIT is covering their butts from a potential lawsuit by taking the videos down. If it was for PR reasons, they would have done nothing at all, as this would have brought the least amount of attention to the issue. They wouldn't bring big negative PR to the...
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    Schools Question About Fellowships/Stipends in Grad School

    Thank you for the speedy reply ParticleGrl. I'm trying to decide whether or not to pursue a PhD, and obviously money is a big factor. It is too bad that you have to practically be in poverty to get a PhD.
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    Schools Question About Fellowships/Stipends in Grad School

    I am under the impression that if you are accepted to a graduate school (PhD) with funding, you receive a stipend. I am planning on applying for as many fellowships as I can that are similar to these: http://scgf.orau.gov/benefits.html http://www.krellinst.org/csgf/ http://www.nsfgrfp.org/...
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    Set Theory Question(inclusion-exclusion principle related)

    Homework Statement An auto insurance has 10,000 policyholders. Each policyholder is classified as: (i) young or old; (ii) male or female; (iii) married or single. Of these policyholders, 3000 are young, 4600 are male, and 7000 are married. The policyholders can also be classi ed as 1320...
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    Post Your Grades for Spring Term

    Electricity and Magnetism ~ A- Macroeconomics ~ A Linear Algebra ~ A Econometrics ~ A- Scientific Computing ~ A
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    Why Doesn't the Squeeze Theorem Apply to lim_{x\rightarrow0}\frac{sin(x)}{x}?

    Thanks for the reply. I suppose that makes sense. If the either of the two "squeezing" functions don't have a limit, it is impossible to say anything about the middle function with the squeeze theorem. The fact that cos(x) < sin(x)/x < 1 for -pi/2 < x < pi/2, was the result of the geometric...
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    Why Doesn't the Squeeze Theorem Apply to lim_{x\rightarrow0}\frac{sin(x)}{x}?

    I just read through a geometric proof that made complete sense to me and resulted in: lim_{x\rightarrow0}\frac{sin(x)}{x} = 1 But this contradicts what I thought I knew about the squeeze/sandwich principle. -1 < sin(x) < 1 (-1/x) < [sin(x)/x] < (1/x) The limits at (-1/x) and (1/x)...
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