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Natural Science Forum / Physics / General Physics / March 2007



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Pi with bad calculator

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BioFreak - 19 Mar 2007 01:45 GMT
While vacationing in the Ozarks you're stuck with a
little piece of grocery store calculator whose
extravagance doesn't go beyond a mere square root
button. But you're doing a back of the envelope
calculation of something that begs for a value for Pi
way better than 3.1415 which you can remember for sure.
Other than cursing yourself for not remembering the
next few digits what else can you do to manage?

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Eric Gisse - 19 Mar 2007 02:21 GMT
> While vacationing in the Ozarks you're stuck with a
> little piece of grocery store calculator whose
[quoted text clipped - 4 lines]
> Other than cursing yourself for not remembering the
> next few digits what else can you do to manage?

355/113

> --
>
[quoted text clipped - 8 lines]
>
>                           - Ali Shari'ati
BioFreak - 19 Mar 2007 02:47 GMT
>> While vacationing in the Ozarks you're stuck with a
>> little piece of grocery store calculator whose
[quoted text clipped - 6 lines]
>
> 355/113

That gives two more significant figures, which is
great. This didn't happen of course in the ozarks but
in my bed last week :-) Too lazy to get up and go to
the other room. I did curse though.

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Timo Nieminen - 19 Mar 2007 02:31 GMT
> While vacationing in the Ozarks you're stuck with a
> little piece of grocery store calculator whose
[quoted text clipped - 4 lines]
> Other than cursing yourself for not remembering the
> next few digits what else can you do to manage?

Of course, you mean 3.1416 as the 5 digit approximation.

There are the usual mnemonics such as "How I need a drink. Alcoholic, of
course, after the heavy chapters involving quantum mechanics."

Or rational approximations such as 355/113, known for about 1.5 millenia,
allowing you to get 7 digits of precision through memorisation of a mere
6!

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mmeron@cars3.uchicago.edu - 19 Mar 2007 02:54 GMT
>> While vacationing in the Ozarks you're stuck with a
>> little piece of grocery store calculator whose
[quoted text clipped - 13 lines]
>allowing you to get 7 digits of precision through memorisation of a mere
>6!

If needed.  Can't think of many back of the envelope calculations for
which the rule "pi is 3, pi^2 is 10" is not good enough.

Mati Meron                      | "When you argue with a fool,
meron@cars.uchicago.edu         |  chances are he is doing just the same"
BioFreak - 19 Mar 2007 03:52 GMT
> If needed.  Can't think of many back of the envelope calculations for
> which the rule "pi is 3, pi^2 is 10" is not good enough.

Not when both very large and very small numbers are
simultaneously present.

Back of the envelope calculations compromise accuracy.
But it sometimes needs high precision to attain even
that.  

> Mati Meron                      | "When you argue with a fool,
> meron@cars.uchicago.edu         |  chances are he is doing just the same"

"No Lord but Jehovah; no tax but that of the Temple; no
friend but the Zealots."
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BioFreak - 19 Mar 2007 19:09 GMT
> Back of the envelope calculations compromise accuracy.
> But it sometimes needs high precision to attain even
> that.

No it can easily be said the opposite way also. The
first sentence I mean. It depends on where the absence
of precision is applied. To the plan of solution or to
the numbers involved. In what I said it applied to the
plan.  

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   "zertesh ghamsur shod."

BioFreak - 19 Mar 2007 03:23 GMT
>> While vacationing in the Ozarks you're stuck with a
>> little piece of grocery store calculator whose
[quoted text clipped - 6 lines]
>
> Of course, you mean 3.1416 as the 5 digit approximation.

No I had no clue what came after 5 in there.

> There are the usual mnemonics such as "How I need a drink. Alcoholic, of
> course, after the heavy chapters involving quantum mechanics."
>
> Or rational approximations such as 355/113, known for about 1.5 millenia,
> allowing you to get 7 digits of precision through memorisation of a mere
> 6!

Remembering 6 digits, I found out just now, can get you
9 significant figures if you use Ramanujan's
approximation. It does need the square root button.
Calculate 97.5-(1/11), then take the square root of the
result two times :)

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Richard Tobin - 19 Mar 2007 03:27 GMT
>There are the usual mnemonics such as "How I need a drink. Alcoholic, of
>course, after the heavy chapters involving quantum mechanics."

Now I, even I, would celebrate in rhymes unapt
The great immortal Syracusan rivaled nevermore
Who in his wondrous lore passed on before
Gave men his guidance how to circles mensurate.

-- Richard
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Nomen Lapetos - 19 Mar 2007 02:39 GMT
> While vacationing in the Ozarks you're stuck with a
> little piece of grocery store calculator whose
[quoted text clipped - 4 lines]
> Other than cursing yourself for not remembering the
> next few digits what else can you do to manage?

you don't need any more precision than that.

What precision does a slide rule have ?
Phil Holman - 19 Mar 2007 05:12 GMT
> While vacationing in the Ozarks you're stuck with a
> little piece of grocery store calculator whose
[quoted text clipped - 4 lines]
> Other than cursing yourself for not remembering the
> next few digits what else can you do to manage?

Taylor Series ............Pi/4 = 1 - 1/3 + 1/5 - 1/7.........etc

Leibnitz Formula ........ Pi/8 = 1/(1*3) + 1(5*7) + 1(9*11).......etc

Nothing fast here; you have to generate many terms in the series to get
even a moderate accuracy.

Phil H
Sam Wormley - 19 Mar 2007 05:21 GMT
> While vacationing in the Ozarks you're stuck with a
> little piece of grocery store calculator whose
[quoted text clipped - 4 lines]
> Other than cursing yourself for not remembering the
> next few digits what else can you do to manage?

  http://en.wikipedia.org/wiki/Euler's_identity
Jim Black - 19 Mar 2007 05:58 GMT
> While vacationing in the Ozarks you're stuck with a
> little piece of grocery store calculator whose
[quoted text clipped - 4 lines]
> Other than cursing yourself for not remembering the
> next few digits what else can you do to manage?

Well, there are nice algorithms for doing this, but the only one I
remember off the top of my head is

 pi = 4/1 - 4/3 + 4/5 - 4/7 + ...

and that converges way to slow.  And if I could look up an algorithm,
I could look up pi.

So it's back to first principles.  We can use the old trick of
doubling the number of sides in a regular polygon repeatedly to
approximate a circle.  So in my head I draw a little diagram
containing half of a side of an n-gon, and a side of a 2n-gon, and
lines connecting everything to the center.  I set the radius of the
circle to one, and I'll call the length of the half-side of the n-gon
y.  By two applications of the Pythagorean theorem, the length of the
2n-gon side is:

 sqrt( y^2 + (1 - sqrt(1-y^2))^2 )

Simplifying, we get:

 sqrt( 2 - 2 sqrt(1-y^2) )

So what we need to iterate is

 s' = sqrt( 2 - 2 sqrt(1 - s^2/4) ),

starting from s=1 in the case of the hexagon.  But notice that we
would be taking a square root and then immediately squaring it, which
is clearly wasteful.  So let's simplify the operations we have to do
between two of the square roots that can't be eliminated:

 1 - (sqrt(2-2x))^2 / 4

 = (x + 1)/2

That isn't so painful, after all.  First, we do the calculation up to
the square root from s=1, getting

 sqrt(3/4).

Then we iterate

 x' = sqrt((x + 1)/2)

N times, where N is some large number.  Then we finish off the
calculation of the side length with

 s = sqrt(2 - 2x).

Last, we multiply by 3 * 2^(N+1) to get pi.

I tried this for N = 20 on my PC's simulacrum of a stupid calculator.

 I got:  3.1415926535897605995224090804006
 pi =    3.1415926535897932384626433832795

On a real stupid calculator, though, I'd be a bit worried about loss
of precision.  Notice that at the end of the calculation,

 sqrt(2 - 2x) ~ pi / 3 / 2^(N+1)

so

 x ~ 1 - pi^2 / 9 / 2^(2N+3).

That's going to limit how far you can take the calculation out, unless
you want to spend even more time thinking and come up with a clever
way to get around the problem.

--
Jim E. Black
The Ghost In The Machine - 19 Mar 2007 07:18 GMT
In sci.physics, Jim Black
<tramspap@yahoo.com>
wrote
on 18 Mar 2007 21:58:45 -0700
<1174280325.623898.235740@o5g2000hsb.googlegroups.com>:
>> While vacationing in the Ozarks you're stuck with a
>> little piece of grocery store calculator whose
[quoted text clipped - 77 lines]
> --
> Jim E. Black

This isn't too bad.  I might approach this from a slightly
different perspective, by noting that the perimeter is
N times the side, for some value of N, then taking that
chord as an abstract value, and computing the perimeter directly.

So let's assume s is the side of a regular polygon inscribed in the
circle, and p = N * s is our starting estimate for 2*pi.  Therefore,
s/2 is the side of a right triangle whose hypotenuse is 1 (the radius
of a unit circle); the other side is therefore sqrt(1 - s^2/4).
Another right triangle has 1 - sqrt(1 - s^2/4) as one side (this is
the radius between chord and the unit circle), s/2 as the other side,
and the new hypotenuse s'; this new hypotenuse is of course

s' = sqrt((1 - sqrt(1 - s^2/4))^2 + s^2/4)
  = sqrt((1 + 1 - s^2/4 - 2*sqrt(1 - s^2/4)) + s^2/4)
  = sqrt(2 - 2*sqrt(1 - s^2/4))

Since p = N * s and p' = 2 * N * s',  we can multiply both sides
by 2 * N and get:

p' = 2 * N * sqrt(2 - 2*sqrt(1 - s^2/4))
  = 2 * sqrt(2*N^2 - 2*N^2*sqrt(1 - s^2/4))
  = 2 * sqrt(2*N^2 - N*sqrt(4 * N^2 - N^2*s^2))
  = 2 * sqrt(2*N^2 - N*sqrt(4 * N^2 - p^2))

Not the best of expressions -- I was hoping to simplify
N out entirely -- but one can start with N = 6, p = 6
(a regular hexagon) and hope for the best.  I get:

N   p                              pp(N,p)
6   6.000                          6.211657082460498296373572103
12  6.211657082460498296373572103  6.265257226562476394323498940
24  6.265257226562476394323498940  6.278700406093734414270293643
48  6.278700406093734414270293643  6.282063901781019276222705853
96  6.282063901781019276222705853  6.282904944570924150901218618
192 6.282904944570924150901218618  6.283115215823715291032926690
384 6.283115215823715291032926690  6.283167784296636817337939207
...
   6.283185307179586476925286767

I can't say I'm thrilled with this but at least it's
converging about as fast, or faster than, 1/N.  The main
problem with my variant is the outer subtraction, which
will lose precision for large N.

If one takes q = (4*N^2 - p^2), q' = (16*N^2 - p'^2), then

q' = 16*N^2 - (4 * (2*N^2 - N*sqrt(q)))
  = 8*N^2 + 4*N*sqrt(q)

This expression will not lose precision because of subtraction, except
at the final step where we calculate p = sqrt(4*N^2 - q).   However,
q starts out rather large -- for N = 6, q = 108 -- and gets larger.
A non-exponential calculator might have problems. :-)

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Uncle Al - 19 Mar 2007 22:47 GMT
> While vacationing in the Ozarks you're stuck with a
> little piece of grocery store calculator whose
[quoted text clipped - 4 lines]
> Other than cursing yourself for not remembering the
> next few digits what else can you do to manage?

355/113.  Difference is 0.27 ppm absolute.  Easy to remember...
113355, unless you are an idiot

>                           - Ali Shari'ati

Idiot.

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Thomas Johnson - 21 Mar 2007 02:07 GMT
> > While vacationing in the Ozarks you're stuck with a
> > little piece of grocery store calculator whose
[quoted text clipped - 11 lines]
>
> Idiot.

3.14159--I haven't had any problem remembering that since I was a
sophomore in high school.  Why waste time with 113355 and divide this
part by that part?

Who needs pi to anything beyond 3 on a desert island?  If so, 3.14
should be good enough.
The Ghost In The Machine - 21 Mar 2007 08:45 GMT
In sci.physics, Thomas Johnson
<thomas_johnson00@hotmail.com>
wrote
on 20 Mar 2007 18:07:11 -0700
<1174439231.950267.51310@l77g2000hsb.googlegroups.com>:

>> > While vacationing in the Ozarks you're stuck with a
>> > little piece of grocery store calculator whose
[quoted text clipped - 18 lines]
> Who needs pi to anything beyond 3 on a desert island?  If so, 3.14
> should be good enough.

The Earth's orbit is 1.501 * 10^11 m in radius, give or
take (it's actually elliptical and slowly precesses).
Pi to 25 decimal places would give one an accuracy of less
than the classical radius of a proton.

Anything more than that is probably gravy. :-)

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Ben Newsam - 21 Mar 2007 12:02 GMT
>3.14159--I haven't had any problem remembering that since I was a
>sophomore in high school.  Why waste time with 113355 and divide this
>part by that part?

Indeed. If anyone still has trouble remembering pi, the following
might help, attributed to Harvey L. Carter, professor of History at
Colorado College:

'Tis a favorite project of mine
A new value of pi to assign
I would fix it at 3
For it's simpler, you see,
Than 3.14159
Geert-Jan Uytdewilligen - 22 Mar 2007 17:10 GMT
> > > While vacationing in the Ozarks you're stuck with a
> > > little piece of grocery store calculator whose
[quoted text clipped - 18 lines]
> Who needs pi to anything beyond 3 on a desert island?  If so, 3.14
> should be good enough.

In old egypt they said Pi=3
In the Netherlands we have a poem, and from every word we count the
characters.
"Ook u kunt u zeker vergissen, uw zwakke brein kan altijd verkeerd
beslissen."
Wich results in Pi=3,141592653589
My favourite way of calculating pi is this: take a circle with radius one.
Divide it in n segments. Make triangles of the segments.
Calculate the area of the triangles. Let n go to infinity. A=Pi=lim
(n->infinity) n* tan (180/n)
Does your calculator at least have "tan"?
Mitchell Jones - 31 Mar 2007 04:05 GMT
> > > > While vacationing in the Ozarks you're stuck with a
> > > > little piece of grocery store calculator whose
[quoted text clipped - 29 lines]
> Calculate the area of the triangles. Let n go to infinity. A=Pi=lim
> (n->infinity) n* tan (180/n)

***{Pi is not merely the ratio of the circumference of a circle to its
diameter, but also the ratio of half the circumference to the radius.
Your method uses a semicircle of radius 1, divides it into n/2 polygon
segments, drops a perpendicular from the center of the semicircle to the
midpoint of each segment, and takes the tangent of the resulting central
half-angle. (I am assuming the polygon segments are inscribed within the
semicircle, rather than being outside it, since that gives the most
accurate approximation.) Since the side opposite the angle is half the
length of a polygon segment, there are n of them in the semicircle, and
the angle is therefore 180/n degrees. Unfortunately, the side adjacent
to the angle is shorter than the radius, hence less than 1. Fortunately,
however, the adjacent side approaches 1 as the central half-angle
approaches zero, and so the method works when the tangent is used. But
the reason it works is that tan x approaches sin x as x approaches zero.
Thus a better approximation is:

n sin (180/n)

It is closer to the actual value of pi at all values of n than is n tan
(180/n), though at large values of n (above 10^5 on my calculator) the
differences cease to matter.

For example, if n = 6, then n tan (180/n) = 6 tan (180/6) = 3.464...,
whereas n sin (180/n) = 6 sin (180/6) = 3. Note that 3 is closer to
3.14159... than is 3.464...

Why does the sine work better than the tangent? Because the sine is the
ratio of the opposite side to the hypotenuse, and the hypotenuse is
always exactly equal to 1. The tangent is the opposite side divided by
the adjacent side, and, as noted earlier, the adjacent side is always
less than 1, though it converges on 1 as the central half-angle
approaches 0.

Result: the tangent works, but the sine works better.

This is a great way to get pi on a calculator that has no pi key, but
has trig functions. For example, 10^20 sin (180/10^20) = 3.14159265359,
which is all the significant digits my pocket calculator allows. It has
a pi function but it is no better, giving exactly the same number. (On
my calculator the smallest value of n that gives that result to 12
significant digits is 10^7.)

There are lots of examples where it is useful to be aware that as the
angle approaches zero, the sine and the tangent converge. In many of
them, the tangent may be used when the angle is very small, but the sine
must be used when it is large. Stellar aberration, for example, can just
as well be calculated using the tangent as the sine, in most cases,
because the angles tend to be very small. The aberration of downward
falling raindrops caused by the forward motion of an automobile, on the
other hand, is going to be a large angle, and calculating such things
requires use of the sine.

Very interesting stuff.

--Mitchell Jones}***

> Does your calculator at least have "tan"?

*****************************************************************
If I seem to be ignoring you, consider the possibility
that you are in my killfile. --MJ
BioFreak - 26 Mar 2007 03:29 GMT
>>                           - Ali Shari'ati
>
> Idiot.

Listen Chosen, first time you mistook him for an idiot
it cost you your Island of Stability. Next time it may
cost you your Hellhole of Yitzrael :)


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