Need Some Math Help

[1369]

A buddy just sent me an email with a pic of a check some smartass wrote to Verizon with the amount filled out in some sort of scientific equation. Anyone here willing to post the pic and tell me what the amount is for?

I think in college I was supposed to take this type of math class, but I looked at the requirements, said " on this", went over to George's Bar and drank beer.

[1369]

Nevermind on the help posting, here it is:

[zero signal]

"What now es?"

Well, the sum is about $1, and the exponential looks imaginary, so I'd say about $1.002?

[zero signal]

Or, is there a joke that went over my head in there?

[Shelly]

If that's real, that's pretty funny!

[tlongII]

I think it's about a dollar. However, I'm not sure what the "e" is supposed to represent. I also can't tell what power it is raised to. I can see that pi is part of it, but I can't tell what it's being multiplied by?

[MoSpur]

That is too funny.

[shelshor]

I think it's about a dollar. However, I'm not sure what the "e" is supposed to represent. I also can't tell what power it is raised to. I can see that pi is part of it, but I can't tell what it's being multiplied by?

Tho the details are dimmed by the mists of antiquity, I think "e" is the natural logarithm of 2

[zero signal]

I think it's about a dollar. However, I'm not sure what the "e" is supposed to represent. I also can't tell what power it is raised to. I can see that pi is part of it, but I can't tell what it's being multiplied by?

It's an i, probably for sqrt(-1).

Which is pretty damned funny on a real check.

[Nbadan]

Hint:

e^i(pie) = int (pie) i

[AlamoSpursFan]

"I'll be watching the news...and if anything is vandalized or explodes or catches on fire "x" is gonna equal me kicking your ass!" -- Red Foreman

[baseline bum]

Nevermind on the help posting, here it is:

e^(i*pi) = cos(pi)+i*sin(pi) = -1 + i*0 = -1 (can be proved by taking Taylor Series of e^(ix))

1+1/2+1/4+1/8+... = 1/(1-1/2) = 2

$0.002 + $e^(pi*i) + $(1/2+1/4+1/8+...) = $0.002 + $-1 + $(2-1) = $0.002

Ie, the check is for 2/10th of a cent.

Pretty stupid... if he really wanted to be a prick, he should have put a contour integral on the check.

[TheSanityAnnex]

Randall Patrick Munroe is my hero.

[Nbadan]

e^(i*pi) = cos(pi)+i*sin(pi) = -1 + i*0 = -1 (can be proved by taking Taylor Series of e^(ix))

1+1/2+1/4+1/8+... = 1/(1-1/2) = 2

$0.002 + $e^(pi*i) + $(1/2+1/4+1/8+...) = $0.002 + $-1 + $(2-1) = $0.002

Ie, the check is for 2/10th of a cent.

Pretty stupid... if he really wanted to be a prick, he should have put a contour integral on the check.

Nice...how is it you only have $25 in v-bookie cash again?

[1369]

You sure that shouldn't be two thousandths of one cent?

[baseline bum]

You sure that shouldn't be two hundreths of one cent?

Yes

$0.002 = 0.002 dollar * 100 cents/dollar = 0.2 cents = 2/10 cents = 2/10 of a cent

[1369]

Missed the $ sign.

In my industry (heavy construction) 0.002 would read two thousandths (as in inches).

[2Blonde]

You sure that shouldn't be two thousandths of one cent?

No, $0.02 would be two cents, so $0.002 would be two tenths of a cent.

[2Blonde]

Oops, sorry I didn't see the reply

[Spurminator]

This should clarify things for you guys.
http://media.putfile.com/Verizon-Bad-Math

Originally posted here:
http://www.spurstalk.com/forums/show...hlight=verizon

[Nbadan]

http://media.putfile.com/Verizon-Bad-Math

Classic!

[Johnny_Blaze_47]

Somehow, I've got to work "What now, es?" into my daily routine.

[1Parker1]

@ this check....

[FromWayDowntown]

Somehow, my desire to reduce fractions on my checks doesn't seem quite so rebellious.

[zero signal]

e^(i*pi) = cos(pi)+i*sin(pi) = -1 + i*0 = -1 (can be proved by taking Taylor Series of e^(ix))

Crap, that's Euler's law. I'm a EE major, I'm supposed to know that.