BUY Kytril ONLINE WITHOUT PRESCRIPTION http://elegantcode.com/2008/04/18/getting-on-the-bad-side-of-a-geek/?utm_source=rss&utm_medium=rss&utm_campaign=getting-on-the-bad-side-of-a-geek Tue, 08 May 2012 09:13:00 +0000 hourly 1 http://wordpress.org/?v=3.3.1 BUY Kytril ONLINE WITHOUT PRESCRIPTION http://elegantcode.com/2008/04/18/getting-on-the-bad-side-of-a-geek/comment-page-1/#comment-20031 David Starr Mon, 21 Apr 2008 11:52:52 +0000 http://elegantcode.com/2008/04/18/getting-on-the-bad-side-of-a-geek/#comment-20031 Did this check ever pay out? Did this check ever pay out?

]]> BUY Kytril ONLINE WITHOUT PRESCRIPTION http://elegantcode.com/2008/04/18/getting-on-the-bad-side-of-a-geek/comment-page-1/#comment-19947 Jason Walker Fri, 18 Apr 2008 22:45:29 +0000 http://elegantcode.com/2008/04/18/getting-on-the-bad-side-of-a-geek/#comment-19947 Ok... so this is old, but also, the text below the image is erroneous. e^i(pi) is what the check should read, which is equal to -1. So the check is made out for 2 tenths of a cent. which is a response to the verizon math fiasco - the one where the CSR could not understand there is a difference between 2/10ths of a cent and 2 cents. http://www.verizonmath.com/ Ok… so this is old, but also, the text below the image is erroneous.

e^i(pi) is what the check should read, which is equal to -1.

So the check is made out for 2 tenths of a cent. which is a response to the verizon math fiasco – the one where the CSR could not understand there is a difference between 2/10ths of a cent and 2 cents.

http://www.verizonmath.com/

]]> BUY Kytril ONLINE WITHOUT PRESCRIPTION http://elegantcode.com/2008/04/18/getting-on-the-bad-side-of-a-geek/comment-page-1/#comment-19936 dcarver Fri, 18 Apr 2008 18:27:58 +0000 http://elegantcode.com/2008/04/18/getting-on-the-bad-side-of-a-geek/#comment-19936 The memo is pretty good. :-). The memo is pretty good. :-) .

]]> BUY Kytril ONLINE WITHOUT PRESCRIPTION http://elegantcode.com/2008/04/18/getting-on-the-bad-side-of-a-geek/comment-page-1/#comment-19930 David Fri, 18 Apr 2008 17:32:41 +0000 http://elegantcode.com/2008/04/18/getting-on-the-bad-side-of-a-geek/#comment-19930 I think thats i*pi The last part is an infinite geometric series that converges to 1 (http://en.wikipedia.org/wiki/Geometric_progression) \frac12 \frac14 \frac18 \frac{1}{16} \cdots=\frac{1/2}{1-( 1/2)} = 1. That gives: e ^ (i * pi) 1 = 0 which is Euler's identity (http://en.wikipedia.org/wiki/Euler's_identity) Looks like those Bitches be getting .002 0 = $0.002 I think thats i*pi

The last part is an infinite geometric series that converges to 1 (http://en.wikipedia.org/wiki/Geometric_progression)

\frac12 \frac14 \frac18 \frac{1}{16} \cdots=\frac{1/2}{1-( 1/2)} = 1.

That gives:
e ^ (i * pi) 1 = 0 which is Euler’s identity (http://en.wikipedia.org/wiki/Euler's_identity)

Looks like those Bitches be getting .002 0 = $0.002

]]> BUY Kytril ONLINE WITHOUT PRESCRIPTION http://elegantcode.com/2008/04/18/getting-on-the-bad-side-of-a-geek/comment-page-1/#comment-19928 David Fri, 18 Apr 2008 17:25:56 +0000 http://elegantcode.com/2008/04/18/getting-on-the-bad-side-of-a-geek/#comment-19928 I think that is e^(i*pie) The last part is an infinite geometric series that converges to 1 (http://en.wikipedia.org/wiki/Geometric_progression) \frac12 \frac14 \frac18 \frac{1}{16} \cdots=\frac{1/2}{1-( 1/2)} = 1. That gives: e^{i*pi} 1 = 0 which is Euler's identity (http://en.wikipedia.org/wiki/Euler's_identity) Looks like those Bitches be getting .002 0 = $0.002 I think that is e^(i*pie)

The last part is an infinite geometric series that converges to 1 (http://en.wikipedia.org/wiki/Geometric_progression)

\frac12 \frac14 \frac18 \frac{1}{16} \cdots=\frac{1/2}{1-( 1/2)} = 1.

That gives:
e^{i*pi} 1 = 0 which is Euler’s identity (http://en.wikipedia.org/wiki/Euler's_identity)

Looks like those Bitches be getting .002 0 = $0.002

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