Thursday, April 19, 2012

The Amazing François Viète

François Viète (1540 -1603) was one of the foremost mathematicians of the early Renaissance era.

Born into the French upper middle class, he became a lawyer by profession but his high intelligence allowed him to excel in several fields, including mathematics, astronomy and geography.

During the period when Spain and France were at war Viète broke the Spanish military code allowing the French to read secret dispatches. The Spanish King accused the French of using sorcery against him.

François Viète - great French scholar and mathematician.
(Image from Wikipedia commons)

But it was in mathematics that Viète was to leave his greatest mark. Investigating methods of generating π, the ratio between the circumference of a circle to its radius, he discovered one of the classical formulae of mathematics that today is called Viète’s equation. It was one of the very early formulae involving infinite products and nested square roots.

                      .             2               .              2                  .....   =  π//2     
             √(2)              √(2+√(2))           √(2+√(2+√(2)))                         

Inspired by Viète I've been playing around with nested square roots over the last few years, and the results I've come up with are below. The 5th result contains a generalisation for pth roots, and the 6th is a generalisation of Viète's equation

No proofs are included but I thought the results themselves may be of some interest to a mathematician out there.

1.        If  x(n+1)=√(2+xn)  ,  x0=x, n=0,1,2,3…

            then xn=2cos((cos-1(x/2)/2n)) for |x|≤2
            and lim xn = 2.

            Also, for x>2,


2.        If xn+1=√(2-xn)  ,  x0=x, |x|≤2, n=0,1,2,3…

           then xn=2cos(π//3+(-1/2)n (cos-1(x/2)- π//3))

           and lim xn = 1

3.       If  xn+1= (√2/2)(√(2+xn)+√(2-xn))  ,  x0= x, |x|≤2, n=0,1,2,3…

          then xn=2cos(π//6+(-1/2)n (cos-1(x/2)- π//6))

          and lim xn = √3

4.         1  .           1           .                1             ……….  =  √3      
           √2       √(2-√2))             √(2-√(2-√2)))                       2

5.      If xn+1 = {( xn+√ (xn2-4))/2}1/p + {(xn-√ (xn2-4))/2}1/p

         then xn=2cos((cos-1(x/2)/pn)) for |x|≤2, and

         xn=2cosh((cosh-1(x/2)/pn)) for x>2, 

6.                   .              2               .              2              ..   =  arcos(x/2)  
             √(2+x)      √(2+√(2+x))          √(2+√(2+√(2+x)))         (1-(x/2)2)
              for |x|<2

No comments:

Post a Comment