How are first digits of $\pi$ found?

Question

Since Pi or $\pi$ is an irrational number, its digits do not repeat. And there is no way to actually find out the digits of $\pi$ ($\frac{22}{7}$ is just a rough estimate but it's not accurate). I am talking about accurate digits by either multiplication or division or any other operation on numbers.

Then how are the first digits of $\pi$ found -

3.1415926535897932384626433832795028841971693993...

In fact, more than 100,000 digits of $\pi$ are found (sources - 100,000 digits of $\pi$)

How is that possible? If these digits of $\pi$ are found, then it must be possible to compute $\pi$ with some operations. (I am aware of breaking of circle into infinite pieces method but that doesn't give accurate results.)

How are these digits of $\pi$ found accurately? Can it be possible for a square root of some number to be equal to $\pi$?

Answer 1

You make the assumption there is no way to calculate the digits of $\pi$. That is untrue; there exist many formulas to calculate the digits of $\pi$ (with their own proofs of correctness). One of the simplest (though very slow) formulas is the Leibniz formula for $\pi$.

Answer 2

That a number is irrational means only that it cannot be computed merely by dividing one integer by another. Square roots of integers that are not perfect squares are irrational, but they can be computed by other methods that are just a little more complicated than long division. (These methods come down to addition, subtraction, multiplication, division, and sometimes a limited number of trial-and-error steps.)

Methods of computing the digits of irrational numbers simply find some computable quantity that is known to be very close to the true value of the desired number. If you have a formula that is known to be within the bounds $x \pm 0.0005$, for example, you merely need compute the value of that formula to enough places that you can accurately round it to the nearest 0.001, and then you know for sure the first three digits of $x$ to the right of the decimal point. To compute the first $100,000$ digits of $\pi$ requires you to evaluate a formula known to be within the bounds $\pi \pm (0.5\times10^{-100000}).$ This is difficult but not impossible.

Answer 3

One way the ancient mathematicians used was to compute the perimeter of a regular $n$-gon inscribed in a circle of diameter $1$. If one finds a formula for the perimeter of the inscribed $n$-gon, one has a sequence converging to $\pi$, i.e. for any accuracy you require there exist an $n$ such that this formula is accurate enough. This is not very practical though. In reality these ancient mathematicians such as Zu Chongzhi used iterative methods to compute the first digits of $\pi$.

From a more modern perspective, in analysis one seldom defines $\pi$ as the ratio of the circumference of the circle to its diameter. Here is one way: Let $\exp:\mathbb{C}\to \mathbb{C}$ be a function that satisfies $\exp(x+y)=\exp(x)\exp(y)$ and the derivative at $0$ being $1$. Then one can prove that such a function is unique and is realized by the function $\exp(z)=\displaystyle\sum_{n=0}^\infty \frac{z^n}{n!}$. One then defines $\sin(z)=\displaystyle\frac{\exp(iz)-\exp(-iz)}{2i}$. Consider its restriction on $\mathbb{R}$, then $\pi$ is defined as the smallest positive real number $x$ such that $sin(x)=0$. Then we immediately have very strong tools to compute $\pi$.

Answer 4

And there is no way to actually find out the digits of Pi by either multiplication or division or any other operation on numbers.

You're missing some key words there: Pi cannot be found out by any finite number of operations. Pi is typically computed through an infinite series of operations that is known to converge on the actual value. The earliest known series was the perimeter of n-gons; various other series have been found that are easier to compute or converge faster, of which I believe the Chudnovsky series is currently the fastest.

Answer 5

Use any formula for $\pi$ and implement it numerically. An example is $$\pi=4\arctan 1=4\int_0^1{dx\over 1+x^2}\doteq{4\over N}\left({3\over4}+\sum_{k=1}^{N-1}{1\over 1+(k/N)^2}\right)=:p_N\ .$$ Here we have approximated the integral by a trapezoidal sum. Doing the calculations one finds, e.g., $p_{100}=3.14157598692313$.

If you want millions of decimal digits for $\pi$ you of course have to resort to much deeper facts about $\pi$, which then will lead to faster convergence.

Answer 6

I am aware of breaking of circle into infinite pieces method but that doesn't give accurate results

Says who? You, of course, can't break a circle into infinite pieces, but if you break it into a sufficiently large finite number of pieces, you will get some number of accurate digits.

For another example, doing the taylor series for $4 \cdot \arctan(1)$:

$ 4 \cdot \left ( 1 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \cdots \right ) $

You have to do thousands of terms for each decimal digit (for terms i=9997 and 9999 I got ~3.1418 and ~3.1414 respectively), but since it converges you can be confident that the digits which have stopped changing "have been calculated".

Answer 7

One interesting property of $\pi$ is this (though it is very hard to prove):
\begin{align} 4>&\pi\\ &\pi>4-\frac43\\ 4-\frac43+\frac45>&\pi\\ &\pi>4-\frac43+\frac45-\frac47\\ \end{align} etc.

Theoretically, you can use this to get $\pi$ to any number of digits. However, you need to add up hundreds of terms just to prove that $3.14<\pi<3.15$ (or, what is the same, that the first three digits of pi are $3.14$).

Answer 8

In the 18th century, Leonard Euler discovered an elegant formula:

$$\frac{π^4}{90}=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\dots$$

The more terms you add, the more accurate the calculation of $π$ gets.

$$\frac{π^4}{90}=\frac{1}{1^4}=1.000$$

then $π=3.080$

$$\frac{π^4}{90}=\frac{1}{1^4}+\frac{1}{2^4}=1.0625$$

then $π=3.080$

$$\frac{π^4}{90}=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}=1.0748$$

then $π=3.136$

$$\frac{π^4}{90}=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}=1.0788$$

then $π=3.139$ $$\frac{π^4}{90}=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}=1.0804$$

then $π=3.140$

etc

This is a slow way to calculate them, since after the 100th term, $π$ is $3.141592$, but it calculates them nontheless.

Answer 9

Can it be possible for a square root of some number to be equal to π?

We have the following series for $\pi^2$

$$\pi^2=10-\sum_{k=0}^\infty\frac{1}{((k+1)(k+2))^3}=10-\frac{1}{8}-\frac{1}{216}-\frac{1}{1728}-\frac{1}{8000}-\frac{1}{27000}-...$$

(see https://math.stackexchange.com/a/1644137/134791)

Therefore, we can compute $\pi$ to the desired accuracy as the square root of a fraction.

$$\pi \approx \sqrt{10-\sum_{k=0}^m\frac{1}{((k+1)(k+2))^3}}$$