Elegant Code » Getting on the Bad Side of a Geek

David, on April 18th, 2008 at 11:25 am Said:

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
David, on April 18th, 2008 at 11:32 am Said:

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
dcarver, on April 18th, 2008 at 12:27 pm Said:

The memo is pretty good. :-).
Jason Walker, on April 18th, 2008 at 4:45 pm Said:

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/
David Starr, on April 21st, 2008 at 5:52 am Said:

Did this check ever pay out?

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