Comments for Solve This Math Problem, Verizon

Solve This Math Problem, Verizon

2006-12-13T04:22:58Z

Considering their ability to convert cents to dollars, Verizon should have no trouble solving this equation.

Since we're more electrorock than mathcore, we don't know what the equation means, though we suspect it relates to dimensional analysis. — BEN POPKEN

Verizon Math [xkcd] (Thanks to Dan!)

UPDATE: The end result of the equation is... $.002. The second factor turns into -1 +1, and the last one cancels itself out.

Comment from Tina Michie on 2009-02-01

2009-02-02T02:38:27Z

er i ment e 2 pi = 535.4916 :P

Comment from Tina Michie on 2009-02-01

2009-02-02T02:37:47Z

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

Comment from tmweber on 2006-12-15

2006-12-15T06:53:46Z

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

Comment from Mr. Gunn on 2006-12-14

2006-12-14T21:55:21Z

xkcd rocks. That is all.

Comment from con40dmitri on 2006-12-14

2006-12-14T19:43:45Z

hah - that is awesome. Verizon is unbelievable.

Comment from Angiol on 2006-12-14

2006-12-14T19:34:35Z

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

Comment from Nelsormensch on 2006-12-14

2006-12-14T13:12:06Z

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

Comment from cbearnm on 2006-12-13

2006-12-14T01:50:32Z

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?

Comment from Echodork on 2006-12-13

2006-12-14T00:38:59Z

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?

Comment from lukos on 2006-12-13

2006-12-13T21:22:48Z

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.

Comment from acambras on 2006-12-13

2006-12-13T20:49:17Z

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

Comment from lukos on 2006-12-13

2006-12-13T20:48:40Z

cos(pi) = -1 sorry :P

Comment from lukos on 2006-12-13

2006-12-13T20:47:02Z

"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.

Comment from Rectilinear Propagation on 2006-12-13

2006-12-13T18:58:28Z

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

Comment from lukos on 2006-12-13

2006-12-13T18:16:03Z

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.

Comment from SirNuke on 2006-12-13

2006-12-13T07:53:46Z

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.

Comment from Ben Popken on 2006-12-13

2006-12-13T07:13:29Z

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

Comment from wyldhoney on 2006-12-13

2006-12-13T06:23:38Z

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.

Comment from drtimhill on 2006-12-13

2006-12-13T06:12:20Z

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

Comment from ElizabethD on 2006-12-13

2006-12-13T06:05:15Z

Loves it.

Comment from homerjay on 2006-12-13

2006-12-13T05:05:10Z

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!

Comment from pacifistviking on 2006-12-12

2006-12-13T04:55:15Z

I hope that's legal. Awesome.

Comment from acambras on 2006-12-12

2006-12-13T04:25:10Z

Oh, that is HILARIOUS!