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Should Pi be 2 Pi?

  1. Sep 10, 2006 #1
    We know that pi is circumference over the diameter. But how often do we talk about the diameter in any type of math analysis?

    Radius is used 99% of the time so I think it should be more appropriate to define pi as circumference over the radius. It would only differ by a factor of two.

    pi = 6.2831....

    It just seems more right, and some equations might look even more elegant.
     
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  3. Sep 10, 2006 #2

    Gokul43201

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    Not the famous special case of the Euler Relation.
     
  4. Sep 10, 2006 #3
    Engineers use Diameter, not Radius because were smarter. :biggrin:
     
  5. Sep 11, 2006 #4

    Integral

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    This is a classic.

    I are college edumcated. :rofl:
     
  6. Sep 11, 2006 #5
    oh yea it would ruin Euler's Formula. Something so nice combining 5 best math elements in their simplest form is enough to keep Pi...Pi.

    I never studied that formula, I have no idea what it's used for, but I have seen it and it's quite...impressive. I feel sorry for Euler's wife. :)
     
  7. Sep 11, 2006 #6

    Alkatran

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    Oh please, Euler's formula would be arguably more elegant with a division by two in it. It adds another basic operation and important number to the mix.

    Oh, what's that? Elegance isn't important enough to justify re-writing so much math it would take years?
     
  8. Sep 11, 2006 #7
    Well...you can treat that 2 as a square root...I mean look at this:

    [tex]\sqrt{e^{i\pi}}+1=0[/tex] It looks a lot more...like something to scare those that haven't encountered it yet! high-tech :surprised.

    but then...some smart guy will try to make this:

    [tex]\sqrt{e^{i\pi}}+1=0 <=>
    \sqrt{e^{i\pi}}=-1 <=>
    e^{i\pi}=(-1)^{2} <=>
    e^{i\pi}=1 <=>
    e^{i}=1^{\frac{1} {\pi}} <=>
    e^{i}=1[/tex]

    Which raises my question...and I'd really love an answer to this one...

    does [tex] 1^{\pi}[/tex] really equal 1? I mean [tex]\pi[/tex] can't be written as a fraction made of natural numbers due to its irrationality...so how does 1 get raised to it?
     
    Last edited: Sep 11, 2006
  9. Sep 11, 2006 #8

    TD

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    Yes, 1^x = 1 for all real x, not just for rational (or integer, natural) x.
    a^b = e^(b*ln(a)) => 1^pi = e^(pi*ln(1)) = e^(pi*0) = e^0 = 1.
     
  10. Sep 11, 2006 #9
    :bugeye: I never saw the [tex]e^{b*ln(a)}[/tex] part but it's defently handy! Thanks for the answer. So then you'd really have just [tex]e^{i}=1[/tex]? if Pi would be replaced by a Pi/2 in Euler's orriginal formula?
     
  11. Sep 11, 2006 #10

    HallsofIvy

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    Take a look at a tree trunk or column. Which is easier (for an ancient Greek, say) to measure, diameter or radius?
     
  12. Sep 11, 2006 #11
    Very good point. The bored guys in rainy days who were playing in the sand drawing circles didn't use "AutoCAD 2004" but a string and a piece of chalk or a hard object or...whatever. So they put one end of string in the desired center, let the string be the radius and rotated it 360 degrees around the point that they chose. They did not use diameter for anything except maybe to see if the apple basket fits or not in the back of the cart...
     
  13. Sep 11, 2006 #12
    Pi = 2Pi wouldn't ruin euler's formula at all.

    instead e^(i *pi) = -1

    we would have e^(i *pi) = 1

    That's not the main point. Basically pi would be equavilan to 360 degrees.

    so pi/2 would be 180 and so on.

    When you get to polar or spherical coordinates and are faced with integration, it would go alot easier if pi = 2 pi. Residue theorem is based on the radii. Even the definition of circle is based of radius,

    "all points equal distant from a single point, we call that distance the radius"

    Can anyone define circle in term of the diameter?

    Not that pi = 2 pi wouldn't make any difference, it's still technincally correct.
     
    Last edited: Sep 11, 2006
  14. Sep 11, 2006 #13

    radou

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    If it really seemed more right, someone would have thought of it already.
     
  15. Sep 11, 2006 #14
    [tex]e^{i\pi}=1[/tex] can be rewritten as [tex](e^{i})^{\pi}=1[/tex] still...and if you move the [tex]^\pi[/tex] it would just cancel in the 1. So you'd have two formulas...

    [tex]e^{i}=1[/tex] and [tex]e^{i*\pi}=1[/tex] and [tex](e^{i})^{\pi}=1[/tex] and then...well i can already see people trying to set them equal with eachother and whining about how they know when to use one or another or why they're the same etc.

    But the main point is, 0 is prbably one of biggest numerical achievements in mathematics. an equation combining things like e, i and [tex]\pi[/tex] simply...diserves to have a 0 in it in my oppinion...and -1 isn't that charming. 1 is an important number also...coefficients of 1, multiplying fractinos by 1...rewriting 1 in 100 different top/bottom ways...in trigonometry sin^2+cos^2...number 1 is pretty big also. -1 is not cool enough to make it in Euler's :D
     
  16. Sep 11, 2006 #15
    Ah, but you can't just move the [tex]\pi[/tex] over. It is not true in general that [tex]e^{zw}=(e^z)^w[/tex] for any complex numbers [tex]z[/tex] and [tex]w[/tex]. Thus, [tex]e^{\pi i}=1[/tex] implies that [tex](e^{\pi i})^{\frac{1}{\pi}}=1^{\frac{1}{\pi}}=1[/tex], but this does not imply that [tex]e^{\pi i \frac{1}{\pi}}=1[/tex]. In fact, [tex]e^i \approx 0.540302 + 0.841471 i[/tex].
     
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