Oh, that is HILARIOUS!

I hope that's legal. Awesome.

This could only have been funnier if he had actually written out that equasion in words. It would have probably required that he figure out how to write in .5 point text, but would have been a hoot to look at!

Loves it.

Actually, the math is quite funny. The first term (e^(i*pi)) is just part of Euler's equation, which is (e^(i * pi) + 1 = 0) where i is the square root of -1. So in fact (e^(i*pi) is just -1. The second part is also a simple integration that equals +1, so the result is .002 -1 + 1, which is .002. In other words, the author is saying "here is my .002c worth". hehe.

--Tim

He's paying them $.002. e^((i)(pi) = -1, and that sum at the end sums up to 1. [1/2 + 1/4+ 1/8 + 1/16 + ...]

Not like they'd know that or anything.

Turns out it's .002 -1 + 1 = $.002 = two tenths of one cent.

Two tenths of a cent plus e to the pi times square root of negative one plus the sum of one divided by two to the n where n is all integers from one to infinity... Dollars

I haven't the faintest idea why e^(i*pi) equals -1.

Euler's formula is used a lot differential equations. if you haven't taking higher then second year calculus it wont make any since but it's a identity used to solve complex integrals like cosxe^x dx. use the identity (mathematical genius figured this out) e^ix=cosx + isinx into that equation (call it the integral of f(x)) where the 'Real' part of the integral (e^ix)(e^x) dx equals the integral of f(x). Much easier to solve :D. (cos is real and sin is imaginary). Make since? Probably not but that's Euler's.

I'm going to 'show my work' on the next check I write. That is too cool!

"I haven't the faintest idea why e^(i*pi) equals -1."

Euler's formula is used a lot differential equations. if you haven't taking higher then second year calculus it wont make any since but it's a identity used to solve complex integrals like cosxe^x dx. Use the identity (mathematical genius figured this out) e^ix=cosx + isinx into that equation (call it the integral of f(x)) where the 'Real' part of the integral (e^ix)(e^x) dx equals the integral of f(x). Much easier to solve :D. (cos is real and sin is imaginary). Make since? Probably not but that's Euler's. So e^ipi = cospi + isinpi = 1 + 0. The zero is imaginary number but doesn't matter.

cos(pi) = -1 sorry :P

Damn, lukos, now you took all the fun out of it -- I was going to solve that problem on my lunch hour. ;-)

haha sorry acambras :P and sorry for all the posts, I just made an account and it wasn't posting until I made a new one then they all came at once haha.

There is a Verizon ad ON THIS PAGE as I type this comment. How can Consumerist continue to lambast Verizon in your articles even while you're pimping their service in your advertising blocks?

Whiskey tango foxtrot?

A better method that 'might' have worked would have been to start from the ground up.
If I got 1 kb, you would charge 0.002 CENTS
10 kb - 0.02 CENTS
100 kb - 0.2 CENTS
1000 kb - 2 CENTS
10000 kb - 20 CENTS
so
35000 kb ~ 70 CENTS

The bill should have been slightly more than 70 CENTS, not DOLLARS.

Then I think the light might have gone on.

So, what ended up happening. Did Verizon ever admit their error. Did you cave and send the ~$72. How was it resolved?

George got a full refund: http://verizonmath.blogspot.com/2006/12/response-from-veri...

However, there are apparently other people in the same situation that Verizon hasn't responded to one way or the other: http://verizonmath.blogspot.com/index.html

lukos - Yeah, that seems to be a comon problem here.

hah - that is awesome. Verizon is unbelievable.

xkcd rocks. That is all.

http://en.wikipedia.org/wiki/Euler's_identity

Actually it 2^ (2pi) = 535.4916
and .002+ 5354916= 535.493

and the last infinite sum is 1 since 1/2+ 1/4 +..1/2^n. = 1

535.493+ 1 = 536.50 about :x