# How can I derive pi?

I was wondering, how can I derive pi without fancy math (fancy meaning too complex. I'm not taking out calculus or trig or anything, just not too advance concepts). Can anyone help me out?

mfb
Mentor
Derive what?
You first need a definition of pi. Everything else (what exactly do you want? Its numerical value?) has to start from there.

ProfuselyQuarky
Gold Member
You mean pi in relation to a circle or the digits of pi?

Pi as in how can I find out the actual digits of pi if I never know it. Like how can I find out the ratio of the circle's Circumference to it's diameter

ProfuselyQuarky
Gold Member
Pi as in how can I find out the actual digits of pi if I never know it. Like how can I find out the ratio of the circle's Circumference to it's diameter
That requires computation on software....Maple can do about 10,000 digits instantly.

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NascentOxygen
Staff Emeritus
Pi as in how can I find out the actual digits of pi if I never know it. Like how can I find out the ratio of the circle's Circumference to it's diameter
Take a tiled floor and scatter a handful of nails or satay sticks, mix with lots of patience, and you can experimentally determine a value for Pi by a procedure known as Buffon's Needle experiment. http://www.math.leidenuniv.nl/~hfinkeln/seminarium/stelling_van_Buffon.pdf

Don't expect many places of accuracy at first, but to improve on this, rinse and repeat many times.

And did I mention that you'll need lots of patience?

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DrewD, CalcNerd, ProfuselyQuarky and 1 other person
256bits
Gold Member
Lovely. Used to Play Pick Up Sticks. Never would have guessed Pi was involved.

NascentOxygen
Staff Emeritus
Another demonstration uses what is known as a Monte Carlo technique. Draw a square of some convenient size, and inside the square draw the largest possible circle.

Next, scatter lots of dots at random all over the square; keep a count and call this number S dots. Count how many of those random dots fall inside the circle, call this number C. Then the ratio C/S gives us the ratio of the areas of the two figures, viz., Pi/4.

The more dots you scatter, the more accurate the resultant estimate for Pi. Repeat this multiple times to improve the result even further.

I like this method, because after performing it 7 or 8 times you can apply a law of statistics to produce an estimate of Pi to much better precision than you would expect.

Another demonstration uses what is known as a Monte Carlo technique. Draw a square of some convenient size, and inside the square draw the largest possible circle.

Next, scatter lots of dots at random all over the square; keep a count and call this number S dots. Count how many of those random dots fall inside the circle, call this number C. Then the ratio C/S gives us the ratio of the areas of the two figures, viz., Pi/4.

The more dots you scatter, the more accurate the resultant estimate for Pi. Repeat this multiple times to improve the result even further.

I like this method, because after performing it 7 or 8 times you can apply a law of statistics to produce an estimate of Pi to much better precision than you would expect.

I do this with my students for a fun activity every year. With enough drops you can easily get three of four decimals correct. What is your trick for improving precision?

mfb
Mentor
To expect to get 3.14 correct (<1/100 error on 4-pi), you need about 10,000 points. 3.141 or 3.142? 1 million. A computer can easily do that, but it doesn't sound like fun to do it manually.

NascentOxygen
Staff Emeritus
but it doesn't sound like fun to do it manually.
Which provides an incentive to improve the procedure by reducing the labor involved! You don't need the full common central area, so you can eliminate the circle's inscribed square from the picture. And you don't need 4 areas, all identical---just one will do. This leaves us with a 3-sided figure containing a circle segment: the competing areas to be covered with dots now being roughly equal.

I'm not sure how to prove it mathematically, but I have a notion that when there is not a big difference in the two competing areas then Monte Carlo gives us better accuracy with fewer random dots. I think I decided that by taking the ratio C/S for this modified figure, and doubling it, we have the fractional portion of Pi, i.e., our estimate to ⋅14159

mfb
Mentor
I'm not sure how to prove it mathematically, but I have a notion that when the areas are less dissimilar then Monte Carlo gives us better accuracy with fewer random dots.
I can help, because we have a binomial distribution with variance ##N p q## whereas the central value is Np or Nq. You gain much more from removing areas, however - going to a triangle makes the final formula ##\pi = 2 + 2 \frac{inside}{total}##, so the relative uncertainty now applies to a smaller value (~1.14 instead of 3.14). You could now start to cut away a triangle from the outer part as well, going to ##\pi = 2 + 1.3 \frac{inside}{total}## and so on. Effectively you are approximating pi with the areas itself then, the closer you get with one of the areas the better.

NascentOxygen
To expect to get 3.14 correct (<1/100 error on 4-pi), you need about 10,000 points. 3.141 or 3.142? 1 million. A computer can easily do that, but it doesn't sound like fun to do it manually.

I should correct my earlier comment: I meant that I often have accurate answers. I usually only do this experiment in classes that are not advanced enough to understand variance and estimating errors. Most of the time we have gotten decently good values with about 200-500 points per class.

NascentOxygen
Staff Emeritus
Most of the time we have gotten decently good values with about 200-500 points per class.

mfb
Mentor
Even with 500 points, getting 3.14 is more luck than a good estimate. 3.1 or 3.2 should be very frequent, but still not guaranteed.

Question; wouldn't it be possible to derive the number pi just by knowing that it should be equal to a circle's circumference divided by its diameter? If you could use a ruler or something to get a reasonably accurate measurement of the circumference of an existing circle with a known diameter or radius, couldn't you divide that number by the diameter and get a value for pi to within a decimal or two (or more, depending on how accurate your circumference measurement was)? Maybe you could do this with several different circles and average the pi values, for your estimate to be more accurate?

Even with 500 points, getting 3.14 is more luck than a good estimate. 3.1 or 3.2 should be very frequent, but still not guaranteed.

Yes, it was luck. I didn't mean to encourage people to guarantee good values with only a few hundred points.

mfb
Mentor
Question; wouldn't it be possible to derive the number pi just by knowing that it should be equal to a circle's circumference divided by its diameter? If you could use a ruler or something to get a reasonably accurate measurement of the circumference of an existing circle with a known diameter or radius, couldn't you divide that number by the diameter and get a value for pi to within a decimal or two (or more, depending on how accurate your circumference measurement was)? Maybe you could do this with several different circles and average the pi values, for your estimate to be more accurate?
Sure, you can use a ruler. Other methods are just more accurate.

down to earth
NascentOxygen
Staff Emeritus