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  #21  
Old 01 March 2007, 02:35 AM
S.A. Lowell
 
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This is just terrible. It reminds me of a little math problem I had in elementary school. I would always tell the other students (friends?) that 100*100 was 10,000. But they would always insist that it was 1,000 D=.

But of course, this Verizon thing is much much worse. MUCH.
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  #22  
Old 01 March 2007, 04:57 PM
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I remember hearing the calls a while ago, but I haven't seen the check. Hilarious. And I love the memo line.
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  #23  
Old 13 March 2007, 01:00 AM
an1373
 
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So I'm a few days behind on this, because my dad just forwarded the scan to me and asked me what the amount was which is actually one dollar.

Starting with summation of 1/(2^n). The limit of this summation as n approaches infinity is 1 (really 0.99999999... ...9). You can verify this by using a calculator or writing a for loop (for 1000 terms, matlab returns 1).

For e^(i*pi) when using imaginary numbers in such a notation you can convert it to rectangular form by taking the cosine of the real term and the sine of imaginary coefficient. The sine of pi is 0, so it's just a fancy way of writing 0.

As for the .002, because the limit as n approaches infinity of the summation on the check is technically 0.9999999... ...9, I think he threw that in to round up to a dollar.

I ran this by a couple of friends in grad school to double check my work and they agree with me. I could be wrong but I'm fairly certain this is correct.
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  #24  
Old 13 March 2007, 01:33 AM
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Meaty Pop Meaty Pop is offline
 
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Quote:
Originally Posted by an1373 View Post

For e^(i*pi) when using imaginary numbers in such a notation you can convert it to rectangular form by taking the cosine of the real term and the sine of imaginary coefficient. The sine of pi is 0, so it's just a fancy way of writing 0.
You've got this part a little confused. If you take the cosine of the real part, then you're taking cos(0) = 1. It should be: e^(i*pi) = cos(pi) + i*sin(pi) = -1 + i*0 = -1. This is using Euler's formula mentioned previously.

You would never have a real part in the exponent since it could just be split out. Ex. e^(1 + i*pi) = [e^(1)]*[e^(i*pi)]

Also, I'd like to add i'm not sure you can up with a more graceful way or writing -1. Combining e, i, and pi into such a neat and nice formula. It seems exactly like something a mathematician would put.

Last edited by Meaty Pop; 13 March 2007 at 01:40 AM.
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  #25  
Old 23 May 2007, 03:05 PM
silenthorn
 
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I know this thread died a while ago, but I just stumbled on it, and want to take the opportunity to explain the mathematics.

It's true that Euler's identity states that e^(i*pi) + 1 = 0. Stated another way, e^(i*pi) = -1, but why is this the case? One must look at Euler's formula for the answer.

Euler's formula states that e^(i*x) = cos(x) + i*sin(x).
For the special case x=pi, this becomes:
e^(i*pi) = cos(pi) + i*sin(pi)
e^(i*pi) = -1 + i*0
e^(i*pi) = -1

If x=0, we have e^(i*0) = e^0, so we expect the answer to be 1 through the formula, and this is indeed the case:
e^(i*0) = cos(0) + i*sin(0)
e^(i*0) = 1 + i*0
e^(i*0) = 1

In the above special cases, the answer was real (no i). We can also get answers that are imaginary (no real part) when x=pi/2 or x=3*pi/2. Most of the time we get answers that are complex (both real and imaginary parts).



I can attest to the fact that this identity is used in almost every application of electrical engineering (I am an EE student). It is very helpful because it turns trigonometry (sines and cosines) into complex arithmetic. Multiplying two or more sine/cosine functions can get messy rather quickly, but multiplying two exponentials is rather straightforward (just add the exponents). To make life easier, any signal (wave) can be represented as a sum of sine waves. Electrical engineers use these properties to make life a lot easier whenever possible.

That's not to say that Mr. Munroe must be an electrical engineer. I've seen this used in quantum physics, to express probability functions (also waves). Physicists and EE's also share the field of electromagnetics (no pun intended), where this notation is used extensively. I can't claim to know what sort of work a mathematician would use this for, but I'm certain they are all familiar with it. I don't know how extensively this is used in other engineering disciplines. I could imagine uses for it when analyzing vibrating beams and the like, but that's not to say there aren't better mathematical tools for that sort of thing. In any case, Electrical engineer is a safe bet for the type of person that would write this formula, but it could have been done by anyone with a strong grounding in mathematics.




As for the infinite summation of 1/(2^n), this is equivalent to 1. Someone argued that it should be expressed as a limit, but this is not necessary, since the upper limit of the summation is infinity. if it was the sum from n=1 to N, then it would be proper to express it as the limit, as N approaches infinity, of the summation from n=1 to N of 1/(2^n).

If you were to write it out, the summation would look like this:
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...

As you add each term, the summation approaches 1:
0.5, 0.75, 0.875, ...

After 4 terms: 0.938
After 7 terms: 0.992
After 10 terms: 0.9990
After 14 terms: 0.99994
After 17 terms: 0.999992
After 20 terms: 0.9999990
After 24 terms: 0.99999994

After an infinite number of terms, there will be an infinite number of nines. 0.999... with an infinite number of nines is equivalent to 1. There are many proofs for this, just search the internet for one that is intuitive enough for you to agree with.



Since e^(i*pi) + the summation = -1 + 1 = 0, the amount of the check is 0.002 dollars. If you work for Verizon, the amount of the check is 0.002 cents.

For more reading:
http://en.wikipedia.org/wiki/Euler%27s_formula
http://en.wikipedia.org/wiki/0.999...
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  #26  
Old 23 May 2007, 03:53 PM
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Spam & Cookies-mmm Spam & Cookies-mmm is offline
 
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Quote:
Originally Posted by silenthorn View Post
I know this thread died a while ago, but
... it took this long to type up the full explanation.

Thanks, and welcome to the board!
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  #27  
Old 23 May 2007, 04:34 PM
Doug4.7
 
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Quote:
Originally Posted by silenthorn View Post
After an infinite number of terms, there will be an infinite number of nines. 0.999... with an infinite number of nines is equivalent to 1. There are many proofs for this, just search the internet for one that is intuitive enough for you to agree with.
The simple proof is ti attempt to find a number between .99999(on and on) and 1.0. As it is you can't, so .9999.... and 1.0 must be "next to" each other. The only way that can be true is if they are the same. You can always find a real number between any two numbers unless they are the same number.

Or something like that.....It's been a few years...
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  #28  
Old 23 May 2007, 04:58 PM
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Dr. Winston O'Boogie Dr. Winston O'Boogie is offline
 
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Quote:
Originally Posted by Doug4.7 View Post
The simple proof is ti attempt to find a number between .99999(on and on) and 1.0. As it is you can't, so .9999.... and 1.0 must be "next to" each other. The only way that can be true is if they are the same. You can always find a real number between any two numbers unless they are the same number.

Or something like that.....It's been a few years...
Another simple proof:

x = .9999999999...
10x = 9.9999999999...
10x - x = 9.999999999...- .9999999999... = 9
9x = 9
x = 1
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  #29  
Old 24 May 2007, 11:12 PM
forceflow15
 
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waffles. posted by accident while doing the math
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  #30  
Old 26 May 2007, 08:48 AM
HerrLip
 
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I've always found the most useful way to derive the solution to a geometric series to be the following (not really sure of a great way to type out the summation here, though):

SUM(n = 0:inf) a^n = 1 + a + a^2 + a^3 + ... = S

S = 1 + aS = 1 + a(1 + a + a^2 + ...) = 1 + a + a^2 + a^3 + ... = S

S = 1 / (1-a)


Not that it's that hard to just memorize the solution, but there's enough crap to memorize. Plus, in a similar fashion, you can easily derive solutions for when the summation starts or ends for different values of n. All that .9999... stuff is great for understanding the limit and all, but you don't want to go apeshit with your calculator at grind time.
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  #31  
Old 02 June 2007, 07:08 PM
bjohn13
 
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Quote:
Originally Posted by Dr. Winston O'Boogie View Post
Another simple proof:

x = .9999999999...
10x = 9.9999999999...
10x - x = 9.999999999...- .9999999999... = 9
9x = 9
x = 1
Or:

1/3 = .333 (repeating)
2/3 = .666 (repeating)
1/3 + 2/3 = 1
.333 (repeating) + .666 (repeating) = .999 (repeating)
Therefore, .999 (repeating) must equal 1.
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  #32  
Old 04 June 2007, 03:34 PM
Dyfsunctional
 
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We could do this all day:

1/9 = 0.111...
2/9 = 0.222...
3/9 = 0.333...
...
7/9 = 0.777...
8/9 = 0.888...
9/9 = 0.999...

And we all know 9/9 is another way of writing "1," so 0.999... = 1.
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  #33  
Old 05 June 2008, 06:23 AM
retired_math_PhD
 
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United States Email going around with picture of the check and absurd comment

Quote:
Originally Posted by Neffti View Post
Even if other snopesters think he won't be winning any prizes for his choice of equation (not something I'm qualified to comment on), but he DOES deserve a prize for patience on the strength of that linked phone call.
I am trying to track down the source of that email going around, of which I got probably half a dozen already from various people, wherein somebody added an absurd interpretation of the exponential formula. The email subject is about an angry engineer. I found a posting of it at

http://www.flickr.com/photos/97477595@N00/2516910055/

Some other posts here already pointed out, and yes, I AM qualified to verify, that it appears to have been NOT an engineer, but rather a mathematician like me, who wrote the check because the symbol j is used by engineers and i is used by mathematicians.

It really bothers me that nobody noticed the totally wrong interpretation of the exponential formula going around in those emails, and the thing keeps getting sent around. Somebody needs to contact the original author to retract the darn thing!

The issue here is: the commenter wrote e^(2*pi) instead of e^(i*pi) and came up with a huge dollar value
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