CUMANIZ MÜBAREK OLSUN

Ekim 19, 2012 tarihinde yayımlandı by

CUMANIZ MÜBAREK OLSUN

Post by Kanaryam Fenerli:

CUMANIZ MÜBAREK OLSUN

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hayırlı günler

Ekim 18, 2012 tarihinde yayımlandı by

hayırlı günler

Post by Kanaryam Fenerli:

hayırlı günler

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What does it look like when I get a thank you?

Ekim 16, 2012 tarihinde yayımlandı by

Answer by Murat Morrison:

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What is a Hexaflexagon?

Ekim 15, 2012 tarihinde yayımlandı by

Answer by Anders Kaseorg:

Vi Hart just started a series of videos on hexaflexagons.

“Hexaflexagons”
http://www.youtube.com/watch?v=VIVIegSt81k
“Hexaflexagons 2”
http://www.youtube.com/watch?v=paQ10POrZh8
http://www.youtube.com/user/Vihart

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Is Randy Powell saying anything in his 2010 TEDxCharlotte talk or is it just total nonsense?

Ekim 15, 2012 tarihinde yayımlandı by

Answer by Nate Waddoups:

Powell would have his audience believe that these ideas are new, and in fact it seems likely that he himself has no idea that they've actually been around for decades.  These crudely conceived ideas were brought to perfection in the 70s by none other than Chrysler, who applied them to transmission design:

http://www.youtube.com/watch?v=MXW0bx_Ooq4
Over the next 10 years, encabulation (as it came to be known) was applied to all manner of products.  In the late 80s, Rockwell employed the technology in their industrial automation control systems, dubbing it the "retro encabulator" because by that time the fundamentals were somewhat dated – though still very well suited to that particular domain.

http://www.youtube.com/watch?v=RXJKdh1KZ0w
And here we are, three decades after the first products were shipped to customers, being told that it's revolutionary?  I guess it's true that if you wait long enough, everything old is new again.

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What are the most interesting equation plots?

Ekim 15, 2012 tarihinde yayımlandı by

Answer by Prathyush Pramod:

Tupper's Self Referential Formula plots a graph of itself !

http://mathworld.wolfram.com/TuppersSelf-ReferentialFormula.html

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If one raised to any finite power is one, how is one raised to infinity an indeterminate form?

Ekim 15, 2012 tarihinde yayımlandı by

Answer by Jay Wacker:

This is how I would think about it.  You're taking the limit
L= \lim_{x\rightarrow 0} f(x)^{g(x)}
with
\lim_{x\rightarrow 0} f(x)\rightarrow1 and \lim_{x\rightarrow 0} g(x)\rightarrow \infty

The limit you're considering is equal to (thanks to the wonders of logarithms)
L = \lim_{x\rightarrow 0} \exp( g(x) \ln f(x)) \equiv \lim _{x\rightarrow 0} \exp(g(x) h(x))
where
h(x) = \ln f(x).

Since \ln 1 =0, \lim_{x\rightarrow 0} h(x) \rightarrow 0
and hence, using h(x) we're back to a more standard indefinite form to evaluate (and then exponentiate) of
\ln L = \lim _{x\rightarrow 0} g(x) h(x)

This also lets you easily evaluate Jonathan Paulson's example
where g(x)= x and h(x) = \ln(1 + 1/x).
Using the Taylor series for h(x) = 1/x - 1/2x^2 +\cdots
we find
\ln L = \lim_{x\rightarrow 0} x ( 1/x - 1/2x^2 +\cdots) = 1
which when we exponentiate gives
L = e^1

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