# How do you know one approximation to $\pi$ is more accurate than another?

I have always been confused regarding the accuracy of $$\pi$$.

In the books which are written on this subject $$\pi$$ , there are references of people and their methods for finding the value of $$\pi$$. Person A found the value of $$\pi$$ and then it says that person B found more accurate value of $$\pi$$ (may be up to $$100$$ decimal places) and then person C found much more accurate value of $$\pi$$ and this goes on.

My question is: How does the mathematician know that the value of $$\pi$$ which he has calculated based on whatever algorithm he applied is more accurate then the value calculated by the previous mathematician?

Imagine a scenario, where the world has just started and there are only $$2$$ mathematicians (A & B) in it. Mathematician A has calculated the value of of $$\pi$$ as $$3.1547$$ (this value is in ancient Chinese text), now the other mathematician says he has calculated a more accurate value, i.e. $$3.1416$$.

My question is How is the Mathematician B so sure that $$3.1416$$ is a more accurate value?

I mean there is no standard with which to compare.

Wikipedia says: The Indian astronomer Aryabhata used a value of $$3.1416$$. Fibonacci in c. 1220 computed $$3.1418$$. Italian author Dante apparently employed the value $$3.14142$$.

When there is no standard to compare, how will I know which is the correct value of $$\pi$$?

• Do you know any calculus? – user109879 Dec 28 '14 at 18:43
• If the method is any good, it will include an estimate of how accurate it is. I don't know about early methods, but modern methods usually do. Part of the method is in fact providing that estimate. – Raskolnikov Dec 28 '14 at 18:44
• accuracy is measured using the distance to the actual value, i.e. if $\bar\pi$ is an approximation, then, in order to say how accurate that approximation is, you have to provide an estimate for $|\bar\pi-\pi|$ from above. A common (equivalent) measure is the number of correct digits. – Thomas Dec 28 '14 at 18:46
• @Thomas He's asking how we compare against $\pi$ in the first place, as it assumes some knowledge of $\pi$. – user98602 Dec 28 '14 at 18:50
• Thomas, I think the problem is: how do we know what the actual value is in order to have something to compare to? – Unit Dec 28 '14 at 18:50

## 8 Answers

We can prove that $$\pi=4\arctan 1 =4\sum_{i=1}^\infty \frac{-(-1)^i}{2i-1}=4\left(1-\frac 13+\frac 15-\frac 17+\dots\right)$$ There are series that converge much more quickly, but let's pretend that this is the only series so far known. The alternating series theorem says that the error truncating a series like this is of the sign of and smaller than the first neglected term. A lazy mathematician might compute the first hundred terms, getting $3.13159\frac {+0.02}{-0}$. Another might sum the first thousand terms, which would give a maximum error of $0.002$ and would clearly be better. Other formulas also come with a bound on the error, so we know how bad it can be.

• I am so glad that I got the answer to a question which was lingering in my mind for so long.Its a perfect answer.Thanks a lot Ross. – Farhat Dec 28 '14 at 20:57

An easy to understand method of approximating pi is that of polygonal approximation. As is the case here, we often can't quite state the exact value of pi. However, we can say with certainty that pi lies between some lower bound and upper bound.

If the lower and upper bound are within $10^{-n}$ of each other, then we can state with complete certainty the first $n$ digits of pi, since these digits are shared by the upper and lower bound. The challenge, then, is to get the upper and lower bound as close together (and thus as close to pi) as possible.

One of the most common ways to calculate $\pi$ is through a sum of a series.

A simple one to see is the Taylor Series of $\arctan(x)$ evaluated at $1$ yields $\pi / 4$. (Link for WolframAlpha on this Taylor Series: http://goo.gl/XK5O1i). Its easy to see that the sum of the first $n$ terms of the Taylor series is as an approximation of $\pi$ that is less accurate than the sum of the first $n+1$ terms.

This guarantees that we can continuously improve the quality of our approximation of $\pi$. Now the method I just showed is not the fasted way to approximate $\pi$ (by this I mean the one whose error after $n$ steps is the least), but it is certainly one way that we can guarantee continual improvement of our approximation by increasing the number of terms we are using.

Hopefully this helps!

"I mean there is no standard with which to compare."

This isn't necessarily true; the value of $\pi$ may not have been explicity defined numerically, but it's implied elsewhere- take, for example, the circumference of a circle: it's the product of the circle's diameter and $\pi$. If I were a mathematician thousands of years ago, I'd attempt to find $\pi$ by measuring a string to some accurate degree, and calling this measurement the string's circumference.

From there, I'd place this string into a circle to the best of my ability (I'm not sure how the ancient Romans did it, but I suspect they had ways, since geometry was a strong subject of debate) and then use another string to find the maximum distance from one point to another point on the circle, calling this maximal distance the "diameter." From there, it's a matter of using long division to numerically calculate $\pi$.

After computing $\pi$ accordingly, you could take that value, and try multiplying it by the diameter of another circle made with another string of a different length to see just how "off" you are. So in reality, you don't even need two mathematicians to gauge just how accurate your $\pi$ estimation is.

Today, approximations of such irrational numbers are made using Monte Carlo methods- there's a very famous and easy-to-understand one pertaining to figuring out $\pi$. A little research online will direct you to it. The accuracy is basically constrained to computational power; the greater power we have, the more accurately we can approximate $\pi$.

Finally, consider the mathematical breakthrough in the Taylor series. Using series, we can numerically find the value of many irrational numbers (including $\pi$), and reasonably estimate our accuracy.

• Thanks for pointing my misconception regarding the 'standard'. Its a beautiful explanation. – Farhat Dec 28 '14 at 20:55

Without standards, of course, you can't even agree about what $\pi$ means, or really discuss anything. Language itself is a standard.

Standards can be agreed upon between two people, but it is harder to establish them with a small group of people than with a large group of people, because the more minds, the more clarifying the arguments can be.

It's even harder to keep up standards in the physical sciences, where there are fiscal incentives to muddy the water. Think medicine, or global warming, or chemistry and the environment. Mathematics is rarely brought into the realm of practical ideological arguments, although Cantor's ideas about infinity certainly were one example. Most of the mathematical ideological fights are about the foundations, and what types of arguments are allowed, but those arguments probably don't affect any of the computation of digits of $\pi$.

There is a standard. The number $\pi$ has a very precise definition, say, as half of the perimeter of a circle of radius $1$ —there are many, many other definitions which we can prove to be equivalent to this one, so which one we use is more or less just a matter of taste.

There are proved methods how to calculate the n-th digit of $\pi$ that can be used to compare different computations.

This claims π is rational, which is known to be false.

Proof No.1

( π-3)×3^3= 4 A circle of diameter 1 is inscribed within a square where the side of the square and the diameter of the circle is one and the same.

Now the circumference of the square 4 is divided by 3 diameters and the quotient of 1.33333333…is further divided by 3^2 (3 diameters squared) the result is 0.148148148… Accordingly, the rational pi is 3.148148148…

Applying this quantum to the equation ( π-3)×3^3= 4

  (3.148148148…- 3)×3^3= 4


We can arrive back to 4 which is the perimeter of the square. But if we substitute the irrational pi, then the result is

(3.141592653…- 3)×3^3= 3.823001631…

which is lesser than 4. But as the perimeter of the square is 4, therefore, the side of the square and the diameter of the circle inscribed within that square ought to be 1 each but it is only(3.823001631/4)0.9557504078.

Proof No.2. (Pythagorean) ( π-3)×3^3= 4 = 2^2

We can carve out an equilateral triangle from the Archimedean method of hemming a circle between two regular hexagons to approximate the circumference of the circle. Now we can divide the equilateral triangle into two right angles. It becomes a right angle of 30, 60 and 90 degrees.

It is in the proportion of hypotenuse being 2, adjacent being √3 and the opposite being 1 unit wise, the quantum rational pi could be arrived at.

( π-3)={(hypotenuse^2)/(adjacent^2 )}÷(hypotenuse +opposite)^2

( π-3)= {2^2/√(3^2 )}÷(2+1)^2 = 0.148148148........

Applying this quantum to the equation ( π-3)×3^3=4 = 2^2

(3.148148148…-3)×3^3= 2^2

We can return back to the unit of the hypotenuse. But it is impossible with the pi of 3.141592653…..which is irrational.