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Pi has been solved!

  1. Jan 6, 2012 #1
    At least that's what the gentleman working the check-out at my local produce market said to me. I told him that there are an infinite number of digits in pi and he flatly said no. He went on to say that the newest supercomputer solved all the digits and until now, no computer could because it was too long. Apparently he saw something about it on TV. I nodded politely and said goodbye- there were a dozen people queued behind me.

    I'm not completely sure what he was on about. Have there been any recent discoveries or new methods in the calculation of pi?

    I was wearing my pi t-shirt which prompted his remarks.
    Last edited: Jan 6, 2012
  2. jcsd
  3. Jan 6, 2012 #2


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  4. Jan 6, 2012 #3
    It's possible. He certainly misunderstood whatever he heard about it.
  5. Jan 6, 2012 #4
    I have also heard on science shows before about how calculating pi to vast numbers of decimal places is irrelevant as you only need pi to I think it was 15 d.p. to calculate the area of a planck length radius circle
  6. Jan 6, 2012 #5
    You were at the produce market, so he probably meant 'Pie is solved.'
  7. Jan 6, 2012 #6
    I don't know of any practical reason to do it outside of benchmarking computers.

    3.14 is good enough for any calculation I might do; 3 would probably be all I need. It is interesting though.

    Last edited: Jan 6, 2012
  8. Jan 6, 2012 #7
    Sure he wasn't meant to be singing;

    "Ding Pie is Done, Ding Pie is done ..."
  9. Jan 6, 2012 #8
    Just when you think you've found them all, someone goes and finds a few more. Apparently, the most recent discovery occurred just last October. Perhaps he is referring to this.
    Pi - 5 trillion Digits
    Note the text: October 17, 2011: The record has been improved to 10 trillion digits.
  10. Jan 6, 2012 #9
  11. Jan 6, 2012 #10
    I had the same conversation with my dad on 2 pi (for some reason he's absolutely convinced people in the stone ages already knew the value of pi.)

    I have always wondered too what would be the more natural constant, pi or two pi. Is there a geometric reason to prefer one over the other?

    Ahum, never mind and move on. pi = C/d, should have known that.
    Last edited by a moderator: Jan 6, 2012
  12. Jan 6, 2012 #11
    Actually, quite a few people are fans of 2pi being the more important constant (for reasons having to do with radians and the unit circle).
  13. Jan 6, 2012 #12
    You could rationalize pi by just counting things in pi units: [itex] \pi/3, \pi/2, \pi, 2\pi, 3\pi,........[/itex]

    Of course this makes all the rational numbers irrational.
  14. Jan 6, 2012 #13


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  15. Jan 6, 2012 #14
    Any digit? How about the [itex] 64^{64} b_{16}th = 100^{100} b_{10}th[/itex] digit?
    Last edited: Jan 6, 2012
  16. Jan 6, 2012 #15
    It's the base 16 representation of the number pi, not of the digit.
  17. Jan 6, 2012 #16
  18. Jan 6, 2012 #17


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    I can haz pie????
  19. Jan 7, 2012 #18


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    Turbo cannot have pie. But Turbos can haz 2 pies!
  20. Jan 7, 2012 #19


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    YayZ!!! Need to get brushed up on radians to figure out how to divide them.
  21. Jan 7, 2012 #20
    Now I wonder whether the reverse could be more natural. I.e., not two pi, but 1/pi or d/C.

    The most natural manner of understanding pi is the relation between the diameter and the cirumference, but somehow, I think it might be more natural to choose the largest of the two to base your thinking on.

    I.e., if I have a string, and use that as the diameter, I somehow need to measure the circumference by multiplying it - which is hard. But if I have a string, which I use for the circumference, then determining that portion of it which makes up the diameter should be easy, at least, easier than the opposite route. I.e., you could mark it, and try to approximate it by halving the string a few times, and see how that series evolves.

    No idea how the math works out then, though.

    (On most scientific calculators you can 'look' at the start of the series of 1/pi by taking a large power of two, divide it by pi, and switch to binary. Of course, there's no obvious regularity to it.)
    Last edited by a moderator: Jan 7, 2012
  22. Jan 7, 2012 #21
    Sometimes a switch of representation has huge consequences. The original geometrical proof of Pythagoras' lemma a^2=b^2+c^2 was more than three pages. The algebraic proof is a one-liner. So yes, it sometimes makes sense to switch representation.

    Having said that, I don't think that fuzzing with this constant now will have a lot of effect.
  23. Jan 7, 2012 #22
    That's quite a misrepresentation. The geometric proof of Pythagoras' theorem (not sure why you call it a lemma...) is absolutely necessary and should be given before the geometrical proof.

    Indeed, the geometrical proof starts from the basic axioms which are easily accepted. The algebraic proof tries to abstract the situation.

    What's wrong with the algebraic proof?? Well, it hides the difficult parts and makes you think it is all easy.Ffor one, it starts with the notion of an inproduct. An inproduct is a function which satisfies certain properties. Then we say that two vectors are orthogonal if their inproduct is 0. Why would we say that?? Does this notion of orthogonality coincide with what we expect orthogonality to be??

    Why is it true in [itex]\mathbb{R}^2[/itex] that (x,y) and (u,v) are orthogonal iff xu+yv=0? Is this intuitively true? No, it must be carefully derived from intuitive axioms. Pythagoras' theorem is a major result in this.

    The algebraic situation is an abstraction of the geometrical situation. But proving that the algebraic situation really represents what we want it to represent is pretty hard to do. And saying that Pythagoras' theorem is just a one-liner is a major misrepresentation of the situation.
  24. Jan 7, 2012 #23
    I guess we are talking about different things. The most direct proof I know is where you prove an equality between two manners of determining the area of a square of size a+b, consisting of 4 triangles a by b and one square c by c. I.e., one derives (a+b)^2 = c^2 + 4 * (1/2 ab). I agree it hides a lot, but mostly only the assumptions on how to determine the areas of rectangles/triangles/squares, not inproduct.

    Hmm, I think I see your argument. Not sure if you can't do without the inproduct in it.
    Last edited by a moderator: Jan 7, 2012
  25. Jan 7, 2012 #24
    That specific proof you mention is indeed quite a nice geometrical argument. But it is very annoying to make rigorous :frown:
  26. Jan 8, 2012 #25


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    Just in case, by way of saying hello, probably anyone who comes here doesn't need to be told, but there will never be any termination of that number. If there was, that would mean pi was rational, but it isn't. Perhaps next time you see the guy at the market, you could point that out, just focussing on the irrationality of pi. It's a common enough misunderstanding.

    I've lost count of the number of students I've had who start out by believing that decimal numbers are real numbers, which is why when I introduce them I also use the term decimal fractions and emphasise that all decimals are rational.
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