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Method of Circle Division#

There are two types of circle division: area approximation and circumference approximation.

Liu Hui used a method to approximate the value of pi by making the areas of the circle and the inner and outer polygons approach each other and lie between them.

Of course, there are also many challengers both domestically and internationally who approximate pi by approximating the circumference. The principle is roughly as shown in the following figure:

Archimedes calculated the value of pi to be between 3.1408 and 3.1429 by calculating the perimeter of a 96-gon. In practical life, this level of accuracy is already sufficient, and the precise value of pi is more of a show of strength in most cases.

In the more than two thousand years since then, people have continuously subdivided the circle using this method. By the late 16th century, François Viète of France subdivided Archimedes' 96-gon by a factor of 12 and calculated the perimeter of a regular 393,216-gon.

This record was broken in the 17th century by Dutch mathematician Ludolph van Ceulen, who spent 25 years calculating the perimeter of a regular $2^{62}$-gon.

It can be obtained that:

$2^{62}=4,611,686,018,427,387,907$

Yes, that's right, it is a regular 4,611,686,018,427,387,907-gon. He thus determined pi to 35 decimal places:
$3.14159 26535 89793 23846 26433 83279 50288...$
He was very proud of this achievement, to the extent that this number was engraved on his tombstone. Even today, Germans often refer to this number as "Ludolph's number."

Twenty years later, the record was broken by Christoph Grienberger, pushing it to 38 decimal places.

So far, it has been a plain enumeration and description. Next, the boy mentioned in the title will appear—Sir Isaac Newton.

Isolation at Home#

In 1666, at the age of 23, Newton was isolated at home due to the outbreak of the Black Death (a similar scene). Newton became interested in some simple equations, such as:

$(1+x)^2=1+2x+x^2$
$(1+x)^3=1+3x+3x^2+x^3$
$(1+x)^4=1+4x+6x^2+4x^3+x^4$
$...$
Later, he discovered a shortcut to directly obtain the answer without complex calculations.

If we only look at the coefficients of $x$, they are the numbers in Pascal's Triangle (also known as Yang Hui's Triangle). This triangle is easy to calculate. If a certain row is known, the next row can be obtained by adding the adjacent two numbers.

Eventually, a general formula was obtained, which can calculate the numbers in any row without relying on the numbers in the previous row.
$(1+x)^n=1+nx+\frac{n(n-1)x^2}{2!}+\frac{n(n-1)(n-2)x^3}{3!}+...$
This is the binomial theorem.

In the standard binomial theorem, $n \in \mathbb{Z}^+$, which is easy to understand. This formula calculates the expansion of $(1+x)$ raised to various powers.
But Newton didn't care about the domain. He directly applied the formula and tried to change the exponent to $-1$, that is, $\frac{1}{1+x}$, and expanded it as follows:
$\frac{1}{(1+x)}=1-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+...$
It's amazing that the obtained series expansion is for $\frac{1}{1+x}$.
Regarding whether this equation is correct or not, Newton provided the following proof:
$\frac{1}{1+x}(1+x)=(1-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+...)(1+x)=1$

Therefore, he firmly believed that the binomial theorem can be extended to cases where the exponent is a negative number. Next, he tried to change the exponent to a fraction, such as $(1+x)^{\frac{1}{2}}$, which means taking the square root of $(1+x)$.
$\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3-\frac{5}{128}x^4+...$
This is another infinite series. We can use it to calculate $\sqrt{3}$:
$\sqrt{3}=\sqrt{4-1}=\sqrt{4(1-\frac{1}{4})}=2\sqrt{(1-\frac{1}{4})}$
By substituting $x=\frac{1}{4}$ into $(1+x)^{\frac{1}{2}}$, we can obtain a rapidly converging series expansion, and high precision for $\sqrt{3}$ can be obtained by calculating only a few terms.

Getting to the Point#

Newton was very interested in the case where $n=\frac{1}{2}$ because the equation of the unit circle is $x^2+y^2=1$.

From this, we can obtain the equation of the upper half of the circle as $y=(1-x^2)^{\frac{1}{2}}$.

This is related to the binomial expansion he just studied. By replacing $x$ with $x^2$, he now has a formula related to the circle:
$\sqrt{1-x^2}=1-\frac{1}{2}x^2-\frac{1}{8}x^4-\frac{1}{16}x^6-\frac{5}{128}x^8-...$
How can this be used to calculate pi? Fortunately, Newton invented calculus.

If we integrate the curve over the interval $[0,1]$, we can obtain $\frac{1}{4}$ of the area of the circle, which happens to be the unit circle, that is, $\frac{\pi}{4}$.

It can be easily obtained that:
$\pi=4\left[x-\frac{1}{2}\frac{x^3}{3}-\frac{1}{8}\frac{x^4}{4}-\frac{1}{16}\frac{x^7}{7}-\frac{5}{128}\frac{x^8}{8}-...\right]_0^1$
By substituting $x=2$, we can obtain pi, which can be calculated to any number of decimal places.

Then Newton made further adjustments and integrated over $[0,\frac{1}{2}]$:

Each $x$ is halved, and the convergence speed of the series is multiplied by several times the power of $x^2$.

At this time, the integrated area is a sector and the right-angled triangle below it.

Therefore:
$\frac{\pi}{12}+\frac{\sqrt{3}}{8}=\left[\frac{1}{2}-\frac{1}{2}\frac{1}{3}\left(\frac{1}{2}\right)^3-\frac{1}{8}\frac{1}{5}\left(\frac{1}{2}\right)^5-\frac{1}{16}\frac{1}{7}\left(\frac{1}{2}\right)^7-\frac{5}{128}\frac{1}{9}\left(\frac{1}{2}\right)^9-...\right]$
Rearranging the equation, we have:
${\pi}=12\left[\frac{1}{2}-\frac{1}{2}\frac{1}{3}\left(\frac{1}{2}\right)^3-\frac{1}{8}\frac{1}{5}\left(\frac{1}{2}\right)^5-\frac{1}{16}\frac{1}{7}\left(\frac{1}{2}\right)^7-\frac{5}{128}\frac{1}{9}\left(\frac{1}{2}\right)^9-...-\frac{\sqrt{3}}{8}\right]$
Taking only the first five terms, $\pi=3.14161$, with an error of only two parts in one hundred thousand. To achieve the accuracy of Ludolph, it is only necessary to calculate the first fifty terms of the series, which can be completed in a few days.

It seems a bit anticlimactic at this point, but now it has been hundreds of years since Newton's time, and of course, more advanced algorithms have emerged.

• This article refers to the following references or materials:

Wikipedia - Method of Circle Division (Liu Hui)
Wikipedia - Ludolph van Ceulen
YouTube-Veritasium - The Discovery That Transformed Pi