I am pretty sure it's the first time I write about John Pell, an English 17th century mathematician.
He was born on March 1st, 1611, in Southwick, Sussex. When he was five and six, his father John Pell and mother Mary Holland of Kent died, respectively. However, his \(4+4=8\) kids didn't learn what it means to be an orphan: he only died when he was 74.
During his life, he became an accomplished linguist, a diplomat, and he even proposed a universal language. He wrote about pedagogy, encyclopedism, and pansophy. Because he did all these pretty sophisticated nonmathematical things, I find it pretty weird that we remember him as a mathematician.
And all the mathematical insights he is famous for – Pell's equation and Pell numbers – are related to rational approximations of \(\sqrt{2}\). What he had learned about those things exceeds this blog post just a little bit and a smart 10yearold kid may understand them. The message is that there were times when presenting a few cute mathematical tricks could make you more famous for the following 4+ centuries than learning lots of languages.
Recall that the square root of two, \(\sqrt{2}\), isn't a rational number. Even we can prove it. If it were true that\[
\sqrt 2 = \frac pq, \quad p,q\in \ZZ
\] then we could also see that \[
2 = \frac{p^2}{q^2},\quad p,q\in\ZZ.
\] But that relationship equivalent to \(p^2=2q^2\) is impossible because if you expand \(p^2\) and \(q^2\) as products of primes, the exponents above \(2\) in \(p^2\) and \(q^2\) are even. However, the similar expansion of \(2q^2\) has an odd exponent above \(2\) because we explicitly added one extra factor of \(2\), so the two expressions \(p^2\) and \(2q^2\) can't be equal. Consequently, \(p/q\) and \(\sqrt{2}\) can't be equal, either.
However, you may write fractions \(p/q\) that are approximately equal to \(\sqrt 2\). One approach is Pell's equation. Let us change the notation \(x=p\), \(y=q\), and we want\[
x^2  2y^2 \approx 0
\] more precisely\[
x^2  2y^2 = 1.
\] This is a special case of Pell's equation \(x^2ny^2=1\) for \(n=2\). Because we are only looking for integer solutions \(x,y\in\ZZ\), it is a "Diophantine" equation. Note that the equation describes a hyperbola. Does the hyperbola intersect points with integer coordinates?
Lagrange was able to prove that there are actually infinitely many solutions as long as \(n\neq r^2\), \(r\in \ZZ\). If the number \(n\) were a perfect square, there would be restrictions similar to the restriction that prevented us from writing \(\sqrt 2\) as a fraction in the first place. So generic Diophantine equations that apparently allow "arbitrarily large" solutions for the values of \(x,y\) typically have infinitely many solutions.
The other approach to approximations of \(\sqrt{2}\) that is named after Pell are the Pell numbers – a close cousin of the Fibonacci sequence. Recall that the ratio of adjacent Fibonacci numbers approaches the golden ratio,\[
\varphi = \frac{1+\sqrt 5}{2} \approx 1.618034 \approx 0.618034^{1}
\] Similarly, the square root of two may be approximated as \[
\sqrt 2 = \lim_{n\to\infty} \frac{Q_n}{2P_n}
\] Both sequences \(P_n\) and \(Q_n\) (Pell and PellLucas numbers) may be calculated by Fibonaccilike recursion relation\[
P_n = 2P_{n1}+P_{n2}\\
Q_n = 2Q_{n1}+Q_{n2}
\] i.e. as "twice the previous number plus the number before it". The relationship only differs from the Fibonacci recursion relation by the factor of \(2\). The initial conditions are\[
(P_0,P_1) = (0,1), \quad (Q_0,Q_1) = (2,2).
\] It would be natural to get rid of the factor of \(2\) in \(Q_n\) – which would also make the factor of \(1/2\) disappear in \(\sqrt{2}\approx Q_n / P_n\) while keeping \(Q_n\in \ZZ\) but I hope that the folks had some reason to define \(Q_n\) with the factor of two.
The sequences given recursively yield\[
(P_0,P_1,\dots ) = (0,1,2,5,12,29,70,169,\dots)\\
(Q_0,Q_1,\dots) = (2,2,6,14,34,82,198,478,\dots)
\] and the expressions may be written using binomial formulae that are ultimately derivable from\[
\eq{
P_n &= \frac{ (1+\sqrt 2)^n  (1\sqrt 2)^n }{2\sqrt{2}}\\
Q_n & = (1+\sqrt 2)^n + (1\sqrt 2)^n
}
\] If you substitute these explicit formulae to the recursive relations and the initial choices of \(P_0,P_1,Q_0,Q_1\), you may check that everything works. Because the recursive relation guarantees that the integrality is preserved for all \(P_n,Q_n\), it also shows that the numbers defined by the explicit expressions above are integervalued.
And you may also check that \(Q_n/2P_n\) converges to \(\sqrt{2}\) for large \(n\). If you want to see this fact using the explicit expressions, it is simple: for large \(n\), you may simply neglect the second term in the numerators because it is much smaller than the first one (two geometric series with very different quotients). Only the first terms are left, they are the same, and the only factor by which \(P_n\) and \(Q_n\) differ are the extra \(1/2\sqrt 2\) in the definition of \(P_n\).
The approximation of \(\sqrt{2}\) is very efficient. For example, let's find a fraction approximating the first 22 or 23 digits of \(\sqrt{2}\). Recall that\[
\sqrt{2} \approx 1.41421\,35623\,73095\,04880\,1689
\] This may be approximated e.g. as\[
\frac{Q_{30}}{2P_{30}} \approx 1.41421\,35623\,73095\,04880\,1719
\] You see that 22 digits, and if we properly round them, 23 digits work. That's not bad because the explicit form of the fraction is\[
\frac{Q_{30}/2}{P_{30}} = \frac{ 152,139,002,499 }{ 107,578,520,350 }
\] It's a ratio of two numbers each of which has 12 digits – and it's really about 11.5 digits because they start with the most likely digit, \(1\). To summarize, the fraction uses 23 digits or so, and gives you a 23digit precision in \(\sqrt{2}\), which is pretty efficient. ;)
The political agent John Pell did lots of other pretty cool things. For example, he (or his pupil Johann Heinrich Rahn) introduced obelus, the "÷" sign for division. But to some extent, Pell's contributions may be summarized as some kind of schoolkids' recreational mathematics which is why it is fair that he is not remembered as one of the history's greatest mathematicians.
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