Page 1 of 2 12 LastLast
Results 1 to 10 of 11

Thread: No closer to the end of pi?

  1. Top | #1
    Administrator lpetrich's Avatar
    Join Date
    Jul 2000
    Location
    Lebanon, OR
    Posts
    4,611
    Archived
    16,829
    Total Posts
    21,440
    Rep Power
    74

    No closer to the end of pi?

    yesterday, March 14, was Pi Day, a day that I neglected. So I'm trying to catch up by posting this:

    Even After 31 Trillion Digits, We’re Still No Closer To The End Of Pi | FiveThirtyEight
    On Thursday, Google announced that one of its employees, Emma Haruka Iwao, had found nearly 9 trillion new digits of pi, setting a new record. Humans have now calculated the never-ending number to 31,415,926,535,897 (get it?) — about 31.4 trillion — decimal places. It’s a Pi Day miracle!

    Previously, we published a story about humans’ pursuit of pi’s infinite string of digits. To celebrate Pi Day, and the extra 9 trillion known digits, we’ve updated that story below.
    That announcement: How Emma Haruka Iwao broke the Guinness World Records title for the most accurate value of pi
    While I’ve been busy thinking about which flavor of pie I’m going to enjoy later today, Googler Emma Haruka Iwao has been busy using Google Compute Engine, powered by Google Cloud, to calculate the most accurate value of pi—ever. That’s 31,415,926,535,897 digits, to be exact. Emma used the power of the cloud for the task, making this the first time the cloud has been used for a pi calculation of this magnitude.

    Here’s Emma’s recipe for what started out as a pie-in-the-sky idea to break a Guinness World Records title:

    Step 1: Find inspiration for your calculation.
    When Emma was 12 years old, she became fascinated with pi. ...

    Step 2: Combine your ingredients.
    To calculate pi, Emma used an application called y-cruncher on 25 Google Cloud virtual machines. ...

    Step 3: Bake for four months.
    Emma’s calculation took the virtual machines about 121 days to complete. ...

    Step 4: Share a slice of your achievement.
    Emma thinks there are a lot of mathematical problems out there to solve, and we’re just at the beginning of exploring how cloud computing can play a role. ...
    Pi in the sky: Calculating a record-breaking 31.4 trillion digits of Archimedes’ constant on Google Cloud | Google Cloud Blog

    It is mathematically impossible to find every digit, but it is still an interesting exercise in computing to see how many digits one can find.

  2. Top | #2
    Super Moderator
    Join Date
    Jun 2002
    Location
    Toronto
    Posts
    15,800
    Archived
    42,293
    Total Posts
    58,093
    Rep Power
    83
    I just checked that guy's result and he made a mistake at digit 456,764,235,769, so we can't really draw any conclusions at all from what he calculated after that one.

  3. Top | #3
    Veteran Member
    Join Date
    Nov 2017
    Location
    seattle
    Posts
    4,000
    Rep Power
    10
    There was a ST episode were a malignant entity took control of the computer.

    To drive it out Spock had the computer calculate PI.

  4. Top | #4
    Shrunken Member WAB's Avatar
    Join Date
    Apr 2004
    Location
    Lake Havasu City, AZ
    Posts
    2,187
    Archived
    2,174
    Total Posts
    4,361
    Rep Power
    59
    Calculating pi is easy-peasy. You just cut it into 8 slices, and serve.







    If you want to get laid, go to college. If you want an education, go to a library. - Frank Zappa

  5. Top | #5
    Sapere aude Politesse's Avatar
    Join Date
    Feb 2018
    Location
    Chochenyo Territory, US
    Posts
    1,671
    Rep Power
    6
    Forgive my ignorance, I know little of mathematics beyond what a social science degree affords. And the conclusion of Carl Sagan's book Contact, in which this particular number played what I assume was a fictive role of sorts. So, π and e are the most well-known, but not by any means the only transcendental numbers derived from nature; are there any others that we have invested as much time in trying to calculate out? Is there a purpose to the venture, aside from the numerological iconicity involved?

  6. Top | #6
    Quantum Hot Dog Kharakov's Avatar
    Join Date
    Aug 2000
    Location
    OCCaUSA
    Posts
    4,334
    Archived
    3,383
    Total Posts
    7,717
    Rep Power
    73
    Natural log of #2 turns up frequently. It's sort of cattycorner to pi, although related.

    <br />
\begin{array}{|c|c|c|c|c|c|c|}<br />
\hline   & \frac{\zeta(1)}{1} & \frac{\zeta(2)}{2}& \frac{\zeta(3)}{3}& \frac{\zeta(4)}{4} & \frac{\zeta(5)}{5}&\dots\\<br />
\hline \lim\limits_{x\to 1^+} log(\frac{x}{x-1}) & \frac{1}{1 \cdot 1^1} & \frac{1}{2 \cdot 1^2}  & \frac{1}{3 \cdot 1^3}& \frac{1}{4 \cdot 1^4}& \frac{1}{5 \cdot 1^5}&\dots\\<br />
\hline log (\frac{2}{1}) & \frac{1}{1 \cdot 2^1} & \frac{1}{2 \cdot 2^2}  & \frac{1}{3 \cdot 2^3}& \frac{1}{4 \cdot 2^4}& \frac{1}{5 \cdot 2^5}&\dots\\<br />
\hline log (\frac{3}{2}) & \frac{1}{1 \cdot 3^1} & \frac{1}{2 \cdot 3 ^2}  & \frac{1}{3 \cdot 3^3}& \frac{1}{4 \cdot 3^4}& \frac{1}{5 \cdot 3^5} &\dots\\<br />
\hline log (\frac{4}{3}) & \frac{1}{1 \cdot 4^1} & \frac{1}{2 \cdot 4^2}  & \frac{1}{3 \cdot 4^3}& \frac{1}{4 \cdot 4^4}& \frac{1}{5 \cdot 4^5} &\dots\\<br />
\hline log (\frac{5}{4}) & \frac{1}{1 \cdot 5^1} & \frac{1}{2 \cdot 5^2}  & \frac{1}{3 \cdot 5^3}& \frac{1}{4 \cdot 5^4}& \frac{1}{5 \cdot 5^5} &\dots\\<br />
\hline \dots &\dots  & \dots&\dots &\dots&\dots&\dots\\<br />
\hline<br />
\end{array}<br />



    Log shift function, with x ? N a ? R : [0,1]:
    log (\frac{x}{x-1}) \to log(\frac{x+1}{x})  \,= \, f(x,a) =\, \sum\limits_{n=1}^\infty \frac{1}{n\cdot x^n} -  \frac{a}{n \cdot x^{2n}}

    Shift logs by a=1:

    <br />
\begin{array}{|c|c|c|c|c|c|c|}<br />
\hline   & \frac{\zeta(1)}{1} & -\frac{\zeta(2)}{2}& \frac{\zeta(3)}{3}& -\frac{\zeta(4)}{4} & \frac{\zeta(5)}{5}&\dots\\<br />
\hline log(\frac{2}{1}) & \frac{1}{1 \cdot 1^1} & -\frac{1}{2 \cdot 1^2}  & \frac{1}{3 \cdot 1^3}& -\frac{1}{4 \cdot 1^4}& \frac{1}{5 \cdot 1^5}&\dots\\<br />
\hline log (\frac{3}{2}) & \frac{1}{1 \cdot 2^1} & -\frac{1}{2 \cdot 2^2}  & \frac{1}{3 \cdot 2^3}& -\frac{1}{4 \cdot 2^4}& \frac{1}{5 \cdot 2^5}&\dots\\<br />
\hline log (\frac{4}{3}) & \frac{1}{1 \cdot 3^1} &- \frac{1}{2 \cdot 3 ^2}  & \frac{1}{3 \cdot 3^3}& -\frac{1}{4 \cdot 3^4}& \frac{1}{5 \cdot 3^5} &\dots\\<br />
\hline log (\frac{5}{4}) & \frac{1}{1 \cdot 4^1} & -\frac{1}{2 \cdot 4^2}  & \frac{1}{3 \cdot 4^3}& -\frac{1}{4 \cdot 4^4}& \frac{1}{5 \cdot 4^5} &\dots\\<br />
\hline log (\frac{6}{5}) & \frac{1}{1 \cdot 5^1} & -\frac{1}{2 \cdot 5^2}  & \frac{1}{3 \cdot 5^3}& -\frac{1}{4 \cdot 5^4}& \frac{1}{5 \cdot 5^5} &\dots\\<br />
\hline \dots &\dots  & \dots&\dots &\dots&\dots&\dots\\<br />
\hline<br />
\end{array}<br />

    zeta to eta shift function, s ?N, a ? R: [0,1]:

      \frac{\zeta(s)}{s} \to \frac{\eta(s)}{s} \, = g(s,a) = \,  \sum\limits_{n=1}^\infty \frac{1}{s \cdot n ^ s}  -  \frac{a}{s \cdot (2n)^ s}

    <br />
\begin{array}{|c|c|c|c|c|c|c|}<br />
\hline   & \frac{\eta(1)}{1} & -\frac{\eta(2)}{2}& \frac{\eta(3)}{3}& -\frac{\eta(4)}{4} & \frac{\eta(5)}{5}&\dots\\<br />
\hline log(\frac{2}{1}) & \frac{1}{1 \cdot 1^1} & -\frac{1}{2 \cdot 1^2}  & \frac{1}{3 \cdot 1^3}& -\frac{1}{4 \cdot 1^4}& \frac{1}{5 \cdot 1^5}&\dots\\<br />
\hline - log (\frac{3}{2}) & -\frac{1}{1 \cdot 2^1} & \frac{1}{2 \cdot 2^2}  &- \frac{1}{3 \cdot 2^3}& \frac{1}{4 \cdot 2^4}& -\frac{1}{5 \cdot 2^5}&\dots\\<br />
\hline log (\frac{4}{3}) & \frac{1}{1 \cdot 3^1} &- \frac{1}{2 \cdot 3 ^2}  & \frac{1}{3 \cdot 3^3}& -\frac{1}{4 \cdot 3^4}& \frac{1}{5 \cdot 3^5} &\dots\\<br />
\hline -log (\frac{5}{4}) &- \frac{1}{1 \cdot 4^1} & \frac{1}{2 \cdot 4^2}  &- \frac{1}{3 \cdot 4^3}& \frac{1}{4 \cdot 4^4}& -\frac{1}{5 \cdot 4^5} &\dots\\<br />
\hline log (\frac{6}{5}) & \frac{1}{1 \cdot 5^1} & -\frac{1}{2 \cdot 5^2}  & \frac{1}{3 \cdot 5^3}& -\frac{1}{4 \cdot 5^4}& \frac{1}{5 \cdot 5^5} &\dots\\<br />
\hline \dots &\dots  & \dots&\dots &\dots&\dots&\dots\\<br />
\hline<br />
\end{array}<br />

    log(\frac{2}{1}) -log (\frac{3}{2}) + log (\frac{4}{3}) -log (\frac{5}{4}) +... \, = \,log(\frac{2}{1}) + log (\frac{2}{3}) + log (\frac{4}{3}) +log (\frac{4}{5}) +log (\frac{6}{5}) +...

    From the above, from the Wallis product for pi:
     \sum\limits_{n=1}^\infty  \,-1^{n+1} \, \cdot \frac{\eta{(n)}}{n} \, = \, log(\frac{\pi}{2})

    When we do a zeta to eta conversion, we lose log(2), and are adding different logs together to get log (pi/4), which is log (pi) - 2 log(2):

    <br />
\begin{array}{|c|c|c|c|c|c|c|}<br />
\hline   & 0\cdot \frac{ \zeta(1)}{1} & - \frac{1}{2} \cdot \frac{\zeta(2)}{2} & \frac{3}{4} \cdot \frac{\zeta(3)}{3} & -\frac{7}{8} \cdot \frac {\zeta(4)}{4} & \frac{15}{16} \cdot \frac{\zeta(5)}{5} & \dots\\<br />
\hline -log(\frac{9}{8}) & 0\cdot\frac{1}{1 \cdot 1^1} & - \frac{1}{2} \cdot\frac{1}{2 \cdot 1^2}  &\frac{3}{4} \cdot \frac{1}{3 \cdot 1^3}& -\frac{7}{8} \cdot \frac{1}{4 \cdot 1^4}&\frac{15}{16} \cdot \frac{1}{5 \cdot 1^5}&\dots\\<br />
\hline  -log (\frac{5^2}{24}) & 0\cdot\frac{1}{1 \cdot 2^1} & - \frac{1}{2} \cdot\frac{1}{2 \cdot 2^2}  &\frac{3}{4} \cdot \frac{1}{3 \cdot 2^3}&-\frac{7}{8} \cdot \frac{1}{4 \cdot 2^4}& \frac{15}{16} \cdot \frac{1}{5 \cdot 2^5}&\dots\\<br />
\hline -log (\frac{7^2}{48}) &0\cdot \frac{1}{1 \cdot 3^1} &- \frac{1}{2} \cdot\frac{1}{2 \cdot 3 ^2}  &\frac{3}{4} \cdot \frac{1}{3 \cdot 3^3}& -\frac{7}{8} \cdot \frac{1}{4 \cdot 3^4}& \frac{15}{16} \cdot \frac{1}{5 \cdot 3^5} &\dots\\<br />
\hline -log (\frac{9^2}{80}) &0\cdot \frac{1}{1 \cdot 4^1} &- \frac{1}{2} \cdot \frac{1}{2 \cdot 4^2}  &\frac{3}{4} \cdot \frac{1}{3 \cdot 4^3}&-\frac{7}{8} \cdot  \frac{1}{4 \cdot 4^4}& \frac{15}{16} \cdot \frac{1}{5 \cdot 4^5} &\dots\\<br />
\hline- log (\frac{11^2}{120}) &0\cdot \frac{1}{1 \cdot 5^1} & - \frac{1}{2} \cdot\frac{1}{2 \cdot 5^2}  &\frac{3}{4} \cdot \frac{1}{3 \cdot 5^3}& -\frac{7}{8} \cdot \frac{1}{4 \cdot 5^4}& \frac{15}{16} \cdot \frac{1}{5 \cdot 5^5} &\dots\\<br />
\hline \dots &\dots  & \dots&\dots &\dots&\dots&\dots\\<br />
\hline<br />
\end{array}<br />

    Gets us the interesting identities:

     pi = e^ { 2\, log(2) \, + \sum\limits_{s=2}^\infty \, \frac{\zeta(s)}{s} \cdot (-1)^{s+1} \cdot (1-2^{1-s})

    And the "log shift function" gets us another interesting identity:

      \sum\limits_{s=2}^\infty \, \frac{\zeta(s)}{s} \cdot (-1)^{s+1} \cdot (1-2^{1-s}) = \sum\limits_{s=2}^\infty \, (\frac{\zeta(s)-1}{s})   \cdot (1-2^{1-s})

    Or....

      \sum\limits_{s=2}^\infty \, \frac{\zeta(s)}{s} \cdot (-1)^{s+1} = \sum\limits_{s=2}^\infty \, \frac{\zeta(s)-1}{s}


    Didn't connect it to pi yet.
    Last edited by Kharakov; 03-18-2019 at 09:57 PM.

  7. Top | #7
    Veteran Member skepticalbip's Avatar
    Join Date
    Apr 2004
    Location
    Searching for reality along the long and winding road
    Posts
    4,487
    Archived
    12,976
    Total Posts
    17,463
    Rep Power
    61
    Quote Originally Posted by Politesse View Post
    Forgive my ignorance, I know little of mathematics beyond what a social science degree affords. And the conclusion of Carl Sagan's book Contact, in which this particular number played what I assume was a fictive role of sorts. So, π and e are the most well-known, but not by any means the only transcendental numbers derived from nature; are there any others that we have invested as much time in trying to calculate out? Is there a purpose to the venture, aside from the numerological iconicity involved?
    There is no purpose that I am aware of for calculating pi to so many decimal places other than the nature of mathematicians to enjoy doing such a thing. But there are several natural, dimensionless numbers that turn up in nature that are of interest to physicists... for example:
    https://www.newscientist.com/article...emake-physics/

    we know the answer to life, the universe and everything. It’s not 42 – it’s 1/137.

    This immutable number determines how stars burn, how chemistry happens and even whether atoms exist at all. Physicist Richard Feynman, who knew a thing or two about it, called it “one of the greatest damn mysteries of physics: a magic number that comes to us with no understanding”.

  8. Top | #8
    Administrator lpetrich's Avatar
    Join Date
    Jul 2000
    Location
    Lebanon, OR
    Posts
    4,611
    Archived
    16,829
    Total Posts
    21,440
    Rep Power
    74
    That's the fine-structure constant, and its value is 1 / 137.035 999 139(31) .

    It is e^2./(4*pi) in quantum-mechanical units, where e is the elementary charge.

  9. Top | #9
    Administrator lpetrich's Avatar
    Join Date
    Jul 2000
    Location
    Lebanon, OR
    Posts
    4,611
    Archived
    16,829
    Total Posts
    21,440
    Rep Power
    74
    A lot of people try to find the value of the fine-structure constant, the ratio of electron and proton masses, and other such physical constants by doing lots of arithmetic on lots of input numbers like small integers and pi. This is usually without any theory that might justify those manipulations.

    Furthermore, these values are their zero-interaction-energy values. Due to quantum-mechanical effects, their effective values are altered by interaction energy. Current advances: The fine-structure constant (NIST) notes that while the fine-structure constant has a value close to 1/137 at zero interaction energy, it goes up a little to around 1/128 at the mass of the W particle. So the FSC is not as fundamental as one might at first think. Its value is also determined by electroweak symmetry breaking, making it dependent on some other constants. The electron-proton mass ratio has a more complicated dependence on other constants.

    Here are all the constants in the Standard Model:
    • Gauge-field (EM-like) "charges": 3
    • Strong CP suppression: 1
    • Higgs mass and self-interaction: 2
    • Quark masses: 6
    • Quark mixing angles: 4
    • Charged-lepton masses: 3
    • Neutrino masses: 3
    • Lepton mixing angles: 4
    • Neutrino Majorana phases: 2

    That gives a grand total of 28 parameters.

  10. Top | #10
    Quantum Hot Dog Kharakov's Avatar
    Join Date
    Aug 2000
    Location
    OCCaUSA
    Posts
    4,334
    Archived
    3,383
    Total Posts
    7,717
    Rep Power
    73
    Quote Originally Posted by lpetrich View Post
    It is mathematically impossible to find every digit, but it is still an interesting exercise in computing to see how many digits one can find.
    Um, Pi is 10 in base Pi. Everyone knows that.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •