How many digits of Pi are enough?

Calculation with ASCII Math

In one of our lectures, the professor mentioned a fascinating concept: with just 10 digits of pi, we can accurately calculate the diameter of the universe with the precision of a single atom. Initially, this assertion may seem astonishing, especially considering our ability to compute pi to many more digits (more like trillions). However, it's a testament to the remarkable relationship between the simplicity of pi and the vastness of our observable universe. Despite the universe being the largest entity we can perceive, it's intriguing that such a fundamental constant as pi can provide insights into its dimensions with such modest input precision.

I want to investigate the reality of this info and found the original article written by NASA. It appears the actual required number of digits more like 40 but still it is significantly smaller than what we can calculate these days.

I wanted to verify it myself;

To calculate the error bound in the circumference calculation using 40 digits of pi, we can use the formula for the error in a calculated value based on the uncertainty in the input values. If we assume that the error in using 40 digits of pi is solely due to the approximation of pi itself, we can use the formula for error propagation in multiplication.

Given:

  • Radius of the universe, \( r \), is around \( 10^{27} \) meters.
  • We're using 40 digits of pi, which means our approximation of pi is accurate to within \( 5 \times 10^{-40} \).

The circumference, \( C \), is calculated using the formula \( C = 2 \pi r \).

The error in this calculation can be estimated using the formula for error propagation in multiplication:

\[ \Delta C = \left| \frac{\partial C}{\partial \pi} \right| \Delta \pi \]

Where \( \Delta C \) is the error in the circumference, \( \Delta \pi \) is the error in the value of pi, and \( \frac{\partial C}{\partial \pi} \) is the partial derivative of circumference with respect to pi.

Partial derivative of circumference with respect to pi:

\[ \frac{\partial C}{\partial \pi} = 2r \]

Given:

  • \( r = 10^{27} \) meters
  • \( \Delta \pi = 5 \times 10^{-40} \)

Substituting into the formula:

\[ \Delta C = 2r \Delta \pi = 2 \times 10^{27} \times 5 \times 10^{-40} = 10^{-12} \]

So, the error in the calculation of the circumference using 40 digits of pi is \( 10^{-12} \), which means the calculated value is accurate to within \( 10^{-12} \) of the true value.

Calculation of Minimum Digits of Pi for Atom Precision

We want to calculate the minimum number of digits of pi needed to achieve precision at the scale of an atom.

The formula for the circumference of a circle is:

$$ C = 2\pi r $$

where \( C \) is the circumference, \( r \) is the radius, and \( \pi \) is pi.

To ensure precision at the scale of an atom (\(1 \times 10^{-10}\) meters), we need to make sure the error in our calculation is smaller than the size of an atom.

The error in the circumference calculation (\( \Delta C \)) should be smaller than the size of a hydrogen atom.

Using the error propagation formula:

$$ \Delta C = \left| \frac{\partial C}{\partial \pi} \right| \Delta \pi $$

We want \( \Delta C \) to be smaller than \(1 \times 10^{-10}\) meters.

From the previous calculation, we found that \( \Delta C = 10^{-12} \) when using 40 digits of \( \pi \).

So, to ensure the error is smaller than \(1 \times 10^{-10}\), we need:

$$ \Delta \pi < \frac{1 \times 10^{-10}}{2r} $$

Given \( r = 10^{27} \) meters, we get:

$$ \Delta \pi < \frac{1 \times 10^{-10}}{2 \times 10^{27}} = 5 \times 10^{-38} $$

I think more realistic calculation would be calculating the precision based on available atoms in the obversable universe rather than assuming it is all filled with atoms as only 4.9% of it is ordinary (baryonic) matter.

Anyways I will stop here. It is an interesting information and wanted share it!