Wallisian product

The Valais product , also Valais product , is a product representation of the circle number , that is, it is a product with an infinite number of factors whose limit value is. It was discovered in 1655 by the English mathematician John Wallis . For this he used a checkerboard-like 'interpolation' between the numerical sequences of Pascal's triangle figured (in whole dimensions) to determine 4 / as the mean binomial coefficient between the zeroth and first dimension. In 2015 a connection with quantum mechanical calculations regarding the hydrogen atom was established for the first time. ${\displaystyle \pi }$ ${\displaystyle \pi }$ ${\displaystyle \pi }$

formula

The product is usually presented in the form:

${\displaystyle {\frac {\pi }{2}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot \dots }$

The abbreviated form of the Valais product results from a transformation as follows:

${\displaystyle {\frac {\pi }{2}}=\left({\frac {2}{1}}\cdot {\frac {2}{3}}\right)\cdot \left({\frac {4}{3}}\cdot {\frac {4}{5}}\right)\cdot \left({\frac {6}{5}}\cdot {\frac {6}{7}}\right)\cdot \ \dots =\prod _{i=1}^{\infty }{\frac {(2i)(2i)}{(2i-1)(2i+1)}}=\prod _{i=1}^{\infty }{\frac {4\,i^{2}}{4\,i^{2}-1}}=\prod _{i=1}^{\infty }\left(1+{\frac {1}{4i^{2}-1}}\right)}$

For the reciprocal it follows:

${\displaystyle {\frac {2}{\pi }}=\prod _{i=1}^{\infty }{\bigg (}1-{\frac {1}{4\,i^{2}}}{\bigg )}}$

The convergence of this product follows from the convergence of the infinite series

${\displaystyle \sum _{i=1}^{\infty }{\frac {-1}{4\,i^{2}}}}$ or. ${\displaystyle \sum _{i=1}^{\infty }{\frac {1}{i^{2}}}}$

Convergence speed

N	2 * product	2 * product / pi	relative error
1	2.7	0.85	15%
2	2.8	0.91	9%
3	2.9	0.93	7%
10	3.07	0.976	2.4%
100	3.134	0.9975	0.25%
1000	3.1408	0.99975	0.025%
10,000	3.14151	0.999975	0.0025%
100,000	3.141585	0.9999975	0.00025%
1000000	3.1415918	0.99999975	0.000025%
${\displaystyle \lim _{n\to \infty }}$	3.14159265 ...	1	0%

The formula is not suitable for an efficient calculation of an approximation of Pi. If one calculates the first 5 terms of the Wallische product and doubles the result, one obtains as an approximation for Pi:

${\displaystyle 2\cdot \prod _{i=1}^{5}{\frac {4\cdot i^{2}}{4\cdot i^{2}-1}}\approx 3{,}002}$

With this approximation, not even the first decimal place could be correctly determined.

After multiplying the first 50 terms, the result is a quotient of two 160-digit numbers, which, however, only provides the approximation 3.126 for Pi, i.e. not even correctly specifying 2 decimal places. Since 3.126 / 3.14159 = 0.9950, the relative error is about 0.5%. The speed of convergence is slower than linear.

The adjacent table shows for some selected values ​​of how good the approximation of pi is, which is obtained after multiplying the terms in the Valais product. The table suggests that the error after multiplying the terms is approximately (e.g. after 100 terms: 0.25% = ). ${\displaystyle N}$ ${\displaystyle N}$ ${\displaystyle N}$ ${\displaystyle {\tfrac {25}{N}}\%}$ ${\displaystyle {\tfrac {1}{400}}}$

This can also be proven by the following mathematical consideration: The quotient between the approximation and the desired value is equal to the infinite product

${\displaystyle \prod _{i=N+1}^{\infty }{\bigg (}1-{\frac {1}{4\,i^{2}}}{\bigg )}}$

With the help of the calculation rules for logarithms, the estimation (for small ones ) as well as by approximating an infinite sum by an integral, you can see that this product has approximately the following value: ${\displaystyle \log(1+x)\approx x}$ ${\displaystyle x}$

${\displaystyle \exp \left(\int \limits _{N}^{\infty }\log \left(1-{\frac {1}{4\,x^{2}}}\right)\,dx\right)\approx \exp \left(\int \limits _{N}^{\infty }-{\frac {1}{4\,x^{2}}}\,dx\right)\approx \exp \left(-{\frac {1}{4\,N}}\right)\approx 1-{\frac {1}{4\,N}}}$.

In order for the first two decimal places to be correct, you therefore need an accuracy of approx. 0.3% (3.13 / 3.14 = 0.997), i.e. for 3 decimal places you need 4 decimal places, etc. ${\displaystyle N=60.}$ ${\displaystyle N=600,}$ ${\displaystyle N=6000}$

Evidence sketch

One defines for which the recursion formula applies. In particular, one obtains for the formula . ${\displaystyle C_{n}(x):=\int \limits _{0}^{x}\sin ^{n}(t)dt}$ ${\displaystyle (n+1)C_{n+1}(x)=-\cos(x)\sin ^{n}(x)+nC_{n-1}(x)}$ ${\displaystyle c_{n}:=C_{n}(\pi /2)}$ ${\displaystyle c_{n+1}={\frac {n}{n+1}}c_{n-1}}$

One calculates and . Well , and therefore . ${\displaystyle c_{2m}={\frac {\pi }{2}}{\frac {1}{4^{m}}}{\binom {2m}{m}}}$ ${\displaystyle c_{2m+1}={\frac {4^{m}}{2m+1}}/{\binom {2m}{m}}}$ ${\displaystyle {\frac {n}{n+1}}={\frac {c_{n+1}}{c_{n}}}{\frac {c_{n}}{c_{n-1}}}<{\frac {c_{n+1}}{c_{n}}}<1}$ ${\displaystyle 1<{\frac {c_{2m}}{c_{2m+1}}}=\pi (m+{\frac {1}{2}}){\binom {2m}{m}}^{2}4^{-2m}<1+{\frac {1}{2m}}}$

In particular , from which one obtains the usual formula by squaring. ${\displaystyle {\sqrt {\pi }}=\lim _{m\rightarrow \infty }{\frac {1}{\sqrt {m}}}{\frac {2\cdot 4\cdot \cdot \cdot (2m)}{1\cdot 3\cdot \cdot \cdot (2m-1)}}=\lim _{m\to \infty }{\frac {1}{\sqrt {m}}}{\frac {(2m)!!}{(2m-1)!!}}}$

physics

By Tamar Friedmann, CR Hagen and students at the university. Rochester (USA) discovered an application of this product in 2015 for the calculation of the error of the quantum mechanical variational calculation of the energy eigenstates in the excited hydrogen atom relative to the solution in Bohr's atomic model.

literature

• John Wallis : The arithmetic of infinitesimals (translation from Latin into English with a foreword by Jacqueline A. Stedall). 1st edition. Springer Verlag, Heidelberg / Berlin / New York 2004, ISBN 0-387-20709-0
• Kurt Endl, Wolfgang Luh : Analysis. An integrated representation; Study book for students of mathematics, physics and other natural sciences from the 1st semester. Volume 2. 7th revised edition. Aula-Verlag, Wiesbaden 1989, ISBN 3-89104-455-0 , p. 343.
• Fridtjof Toenniessen: The secret of the transcendent numbers: A slightly different introduction to mathematics . Springer 2009, ISBN 978-3-8274-2274-3 , pp. 321–322 ( excerpt (Google) )

Web links

Wikibooks: Derivation of the Valais product  - learning and teaching materials

• Johan Wästlund: An Elementary Proof of the Wallis Product Formula for Pi . (PDF; 59 kB) MAA Monthly , December 2007
• Eric W. Weisstein : Wallis Formula . In: MathWorld (English).

Individual evidence

1. ↑ alphagalileo.org ( Memento of the original from November 22, 2015 in the Internet Archive ) Info: The archive link was inserted automatically and has not yet been checked. Please check the original and archive link according to the instructions and then remove this notice. 
2. ↑ phys.org
3. ↑ scinexx.de