Pi approximation games

Yes total nerd :)

But who want to use your PY to calculate the circle circumference, with given diameter

???

(snip)

There are at least two projects to redefine the kilogram, such that it isn't dependent on a physical object.

One involves a more accurate determination of Avogadro's number.

-- glen

You might find fun at the opposite game too, have a look:

bye,

--

piergiorgio

That's known as a 'Selleck' or a 'Magnum PI'.

Cheers

--
Syd

This is cool.

I did this by hand recently...

One could also do the set of possible voltage dividers and corresponding opamp gains.

--

John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro   acquisition and simulation

You are living in a state of sin.

Die ganzen Zahlen hat der liebe Gatt gemacht, alles andere ist Menschenwerk.

DTA

Err... well it will give the correct result in a sufficiently curved space-time. I can't help it if people are using it in the wrong universe!

--

John Devereux

See:

According to the above, they go 3/1, 22/7, 333/106, 355/113,

103993/33102, 104348/33215, the next two you note, and then 833719/265381, well, try out the site.

(snip)

The javascript code for the above site, which says no license and free to copy, uses the following to do the computation:

var maxNumerator = getMaxNumerator(d); var d2 = d; var calcD, prevCalcD = NaN; for (var i = 2; i < 1000; i++) { var L2 = Math.floor(d2); numerators[i] = L2 * numerators[i-1] + numerators[i-2]; if (Math.abs(numerators[i]) > maxNumerator) return;

denominators[i] = L2 * denominators[i-1] + denominators[i-2];

calcD = numerators[i] / denominators[i]; if (calcD == prevCalcD) return;

appendFractionsOutput(numerators[i], denominators[i]);

if (calcD == d) return; prevCalcD = calcD; d2 = 1/(d2-L2); }

I haven't gone through it in detail, but you can see that it is pretty fast. Each is generated directly from the previous one, with no loop through successive denominators.

-- glen

You may want to check the spelling on that one. ;-)

--
Stef    (remove caps, dashes and .invalid from e-mail address to reply by mail)

If all the world's a stage, I want to operate the trap door.
		-- Paul Beatty

Avogadro's number is currently: 6.02214129*10^23 entities per mole and has an uncertainty of plus or minus 27 counts in its last two digits.

Another requires the precise definition of Planck's constant which is:

6.62606957*10^?34 and has an uncertainty of plus or minus 29 counts in its last two digits.

I wonder if either or both of these "constants" can be defined as the ratio of integers. :-)

--
Virg Wall

Any uncertainty is an artifact of measurement. It's a *count*, so ...

Avogadro's Number is a natural number - so yes.

-- Les Cargill

Chuckle... That's great...

I was thinking that most of the time I let 1/2pi =3D 1.6/10 or pi =3D 10 x 5/16 which is better than 1%.

George H.

But who needs to approximate Pi? It's marked right there on the C and D scales!

Slide rules are good for arbitrary rational approximations - look for (nearly) aligned tick marks. And the L scale on a 10" works as a ruler.

-- not sent from my Pickett.

(snip, I wrote)

The ongoing project involves weighing and counting the number of atoms in a silicon sphere. When you figure out how many are in a kg, that could be, should be, an integer.

-- glen

pi/4 = 1 - 1/3 + 1/5 - 1/7 + ...

pi/4 = Sum [(-1)^p]/(2p+1) p = 0 to oo

max error =|[((-1)^p)/(2p+1)]-[(-1)^(p+1)]/(2(p+1)+1)|

for instance if I want to calculate pi to the nearest E = +/- 10^-50, the series will have to be carried out p times such that

10^-50 = |1/(2p+1)-1/(2p+3)|

E(2p+3)(2p+1)=(2p+3) - (2p+1)

E(4p^2 + 8p + 3) = 2

4p^2 + 8p + (3 - 2/E) = 0

p = (-2 +/-{4-(3-2/E)}^.5)/2

for E = 10^-50, p = 10^25

for E = 1/2^128 , p = 10^20

g

ttdesign.com

I've recently discovered a program called Wolfram Alpha. If you have an obsessive interest in series expansions, you can spend hours with this program.

Bob

.

That's the British customary units, not English, that's the language you are confusing with the country which is Great Britain. another common mishtake made by americans.

Probably, but there are a lot of sins. Could you be more specific? :)

The cycles and times were of course measured averages, the inputs would have varied. It was just a quick-and-dirty test. This sort of information seems hard to find for real-world chips.

--

John Devereux