## 5 thoughts on “Getting on the Bad Side of a Geek”

1. David says:

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 2. David says: 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

3. dcarver says:

The memo is pretty good. :-).

4. Jason Walker says:

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/

5. David Starr says:

Did this check ever pay out?