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Facebook: Check Written to Verizon Using Summation Notation

  1. Dec 28, 2011 #1
    Good day all,

    I received this picture through the facebook network last night. I took it as valid at first sight, but for some reason it bothered me; in other words, my skepticism kicked in. The author mentions and assumes the limit as "n goes to infinity" which is annotated in the summation, but there is no limit expressed. I've received a butt chewing previously for not including limits in my expressions, so I am a bit critical about it because I want my math to be outstanding. By the way, I'm sure the author meant to say 'the limit as n approaches infinity.'

    I have attached both a picture of the check sent via facebook and a pdf of what I think he actually meant and what the expression on the check looks like to me. FYI - my program script is still really weak, which is why attached the information.

    First and foremost, please do not give me the answer to his expression. Second, Am I on the right track? If not, I would appreciate some direction, but not the answer. Please note, I do have an answer and reasoning behind it. But of course my reasoning may be wrong if my assumptions are wrong.

    I received some strange explanations of an answer on facebook, but of course it's facebook. In fact, one individual seems to think it's an equation. But I digress.

    Your help is very much appreciated. Thanks in advance!

    Doc
     

    Attached Files:

  2. jcsd
  3. Dec 28, 2011 #2
    no, his math is correct. convert e to the i pi to cosine and sine (euler's equation). then take then treat the sum as doing exactly as he says. a sum that goes from 1 to infinity. it converges at a value. this is an oldy but a goodie.
     
  4. Dec 28, 2011 #3
    The limit is implicit in the summation notation. The definition of an infinite sum is $$\sum_{n=0}^\infty a_n = \lim_{N \to \infty} \sum_{n=0}^N a_n \; .$$

    Also, here's the original source: http://xkcd.com/verizon/
     
  5. Dec 28, 2011 #4
    First and foremost, thank you so much for the responses my friends. I do appreciate it. Concerning the math itself, it is a bit above my pay grade at this point in my mathematical studies, but I will charge forward.

    Secondly, I misread the expression on the check. It looked like e^2*pi, not e^i*pi, so that was my first mistake.

    Thirdly, I had never heard of Euler's Equation before so I did some further research and found a really cool site. Thanks for the direction, TxRationalist. Here is the link (I'm sure there are others):

    http://www.songho.ca/math/euler/euler.html

    Since, e^ix, where pi is substituted for x, is equal to cosx + sinx, which is "almost identical" to cosx + isinx, is equal to 1. I understand there's some manipulation and identity work to be done. Although I understand the Trigonometric relationship between cosx + sinx and 1, I don't quite understand it in a complex number or imaginary number form using e, yet. But I'll give it some time to sink in and review the information from the link.

    I will read up on the sigma notation, so that way I understand that part of the expression a bit better. I should have known it was implicit, but that concept is a fairly new to me. In addition, thanks for the original source, spamiam.

    I'll get back to you when I figure out that last part.
     
  6. Dec 28, 2011 #5

    Fredrik

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    e is equal to cos π + i sin π, and that's not equal to 1. I don't understand your "almost identical" comment. The identity ##e^{ix}=\cos x+i\sin x## holds for all real numbers x. The left-hand side is only equal to ##\cos x+\sin x## when ##\sin x=0##.

    To understand the sum, you must understand this: For all real numbers x such that 0<x<1, we have $$1+x+x^2+\cdots+x^n=\frac{(1+x+x^2+\cdots+x^n)(1-x)}{1-x}=\frac{1+x-x+\cdots+x^n-x^n+x^{n+1}}{1-x}=\frac{1}{1-x}+\frac{x^{n+1}}{1-x}$$ and the last term goes to 0 as n→∞.
     
  7. Jan 3, 2012 #6
    Thanks for your posting, Fredrik.
     
  8. Jan 3, 2012 #7

    Moonbear

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    This is why I love PF! I've seen that circulated a few times, and the skeptic in me simply said, "bounced check."
     
  9. Jan 3, 2012 #8

    Curious3141

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    The caption in that image is wrong. I read the cheque as (dollars)[itex]0.002 + e^{i\pi} + \sum_{n=1}^{\infty}{\frac{1}{2^n}}[/itex]. The infinite sum is equal to 1, and the exponential term is a famous result known to be equal to -1, so those cancel out, leaving the measly amount of $0.002, or a fifth of a cent.

    Is a check for less than a cent legal tender? :tongue:
     
  10. Jan 4, 2012 #9
    Thanks for the posting and reply, Curious3141. Originally, when I first read the document, I thought it said, "e^2pi," which would be 535.4916555. But it doesn't say that. It says "e^ipi." I had never seen or heard of this term before. Obviously this makes the answer to the term and ultimately the expression much different.

    Thanks to the help of PF - every person who posted on this thread - I not only see the difference but I understand it. PF is way cool! :cool:
    I am looking forward to taking advantage of this fantastic resource while I am in school.

    Thanks again my friends!

    "Letting the religious right teach ID in schools is like letting the Marines teach poetry in advanced combat training." J. Simon, Ph.D.
     
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