The digits of pi (π) put to music using Mathematica. The digits are mapped as follows: 1→C, 2→D, 3→E, 4→F5, 5→G, 6→A3, 7→B3, 8→C5, 9→D5, 0→C3. This goes out to 314 digits of pi (π) at 120 bpm.
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Two pages from the book "Arithmetica Infinitorum," by John Wallis. In the table on the left page, the square that appears repeatedly denotes 4/pi, or the ratio of the area of a square to the area of the circumscribed circle. Wallis used the table to obtain the inequalities shown at the top of the page on the right that led to his formula. In 1655 the English mathematician John Wallis published a book in which he derived a formula for pi as the product of an infinite series of ratios. Now researchers from the University of Rochester, in a surprise discovery, have found the same formula in quantum mechanical calculations of the energy levels of a hydrogen atom. "We weren't looking for the Wallis formula for pi. It just fell into our laps," said Carl Hagen, a particle physicist at the University of Rochester. Having noticed an intriguing trend in the solutions to a problem set he had developed for students in a class on quantum mechanics, Hagen recruited mathematician Tamar Friedmann and they realized this trend was in fact a manifestation of the Wallis formula for pi. "It was a complete surprise  I jumped up and down when we got the Wallis formula out of equations for the hydrogen atom," said Friedmann. "The special thing is that it brings out a beautiful connection between physics and math. I find it fascinating that a purely mathematical formula from the 17th century characterizes a physical system that was discovered 300 years later." The researchers report their findings in the Journal of Mathematical Physics. In quantum mechanics, a technique called the variational approach can be used to approximate the energy states of quantum systems, like molecules, that can't be solved exactly. Hagen was teaching the technique to his students when he decided to apply it to a realworld object: the hydrogen atom. The hydrogen atom is actually one of the rare quantum mechanical systems whose energy levels can be solved exactly, but by applying the variational approach and then comparing the result to the exact solution, students could calculate the error in the approximation.
When Hagen started solving the problem himself, he immediately noticed a trend. The error of the variational approach was about 15 percent for the ground state of hydrogen, 10 percent for the first excited state, and kept getting smaller as the excited states grew larger. This was unusual, since the variational approach normally only gives good approximations for the lowest energy levels. Hagen recruited Friedmann to take a look at what would happen with increasing energy. They found that the limit of the variational solution approaches the model of hydrogen developed by physicist Niels Bohr in the early 20th century, which depicts the orbits of the electron as perfectly circular. This would be expected from Bohr's correspondence principle, which states that for large radius orbits, the behavior of quantum systems can be described by classical physics. "At the lower energy orbits, the path of the electron is fuzzy and spread out," Hagen explained. "At more excited states, the orbits become more sharply defined and the uncertainty in the radius decreases."
From the formula for the limit of the variational solution as the energy increased, Hagen and Friedmann were able to pull out the Wallis formula for pi.
The theory of quantum mechanics dates back to the early 20th century and the Wallis formula has been around for hundreds of years, but the connection between the two had remained hidden until now. "Nature had kept this secret for the last 80 years," Friedmann said. "I'm glad we revealed it." More information: "Quantum mechanical derivation of the Wallis formula for pi," by Tamar Friedmann and C.R. Hagen, Journal of Mathematical Physics , November 10, 2015. DOI: 10.1063/1.4930800 , http://scitation.aip.org/content/aip/journal/jmp/56/11/10.1063/1.4930800 Provided by: American Institute of Physics With thanks to Martin Krzywinski and Cristian Ilies Vasile  cool visualisers of Pi. Click "Read More" to watch the video!
Here Come The Mummies performed a chant of the first 143 digits of pi for the track "Easy As Pi". These background vocals made their way on to the album "CuriousTease (Volume 1), where this was found. Calculus Humor decided to take this awesome track and make another Pi Day video for the Pi Day of the century! This is a simulation that estimates Pi by randomly dropping sticks onto a set of horizontal lines spaced one stick length apart. It uses the fact that the probability of a randomly dropped stick intersecting one of the lines is equal to $\frac{\pi}{2}$. (NOTE: YOU NEED TO HAVE ADOBE FLASH INSTALLED TO VIEW) This simulation represents the Buffon's needle problem. Buffon's Needle is one of the oldest problems in the field of geometrical probability. It was first stated in 1777. It involves dropping a needle on a lined sheet of paper and determining the probability of the needle crossing one of the lines on the page. The remarkable result is that the probability is directly related to the value of pi. The problem in more mathematical terms is: Let's take the simple case first. In this case, the length of the needle is one unit and the distance between the lines is also one unit. There are two variables, the angle at which the needle falls (θ) and the distance from the center of the needle to the closest line (D). Theta can vary from 0 to 180 degrees and is measured against a line parallel to the lines on the paper. The distance from the center to the closest line can never be more that half the distance between the lines. The image on the left illustrates that. The needle in the picture misses the line. The needle will hit the line if the closest distance to a line (D) is less than or equal to 1/2 times the sine of theta. That is, D ≤ (1/2)sin(θ). How often will this occur? In the graph below, we plot D along the ordinate and (1/2)sin(θ) along the abscissa. The values on or below the curve represent a hit (D ≤ (1/2)sin(θ)). Thus, the probability of a success it the ratio shaded area to the entire rectangle. What is this to value? The shaded portion is found with using the definite integral of (1/2)sin(θ) evaluated from zero to pi. The result is that the shaded portion has a value of 1. The value of the entire rectangle is (1/2)(π) or π/2. So, the probability of a hit is 1/(π/2) or 2/π. That's approximately .6366197. To calculate pi from the needle drops, simply take the number of drops and multiply it by two, then divide by the number of hits, or 2(total drops)/(number of hits) = π (approximately). Resources

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