MHB Any ideas on Vieta's formula....?

[DreamWeaver]

Evening all! (Sun)Do any of you know of a -relatively - elementary approach to Vieta's forumula, without going into hypergeometric functions and associated repeated fractions, etc?

In short, an ideas or suggestions on how to prove that:

$$\sqrt{ \frac{1}{2} } \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} } } \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} } } }\, \cdots = \frac{2}{\pi}$$Thanks! (Hug)

 

[Opalg]

The Wikipedia article http://en.wikipedia.org/wiki/Vi%E8te%27s_formula mentions Viète's original derivation of his formula by calculating areas of polygons inside a circle. The article also mentions a proof using one of Euler's infinite product representations for $\dfrac{\sin x}{x}$.

Incidentally, it seems that this formula is usually known under François Viète's French name. The latinised version Vieta's formula is associated with formulas for symmetric functions of the roots of a polynomial. But both formulas are the work of the same man.

 

[chisigma]

Evening all! (Sun)Do any of you know of a -relatively - elementary approach to Vieta's forumula, without going into hypergeometric functions and associated repeated fractions, etc?

In short, an ideas or suggestions on how to prove that:

$$\sqrt{ \frac{1}{2} } \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} } } \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} } } }\, \cdots = \frac{2}{\pi}$$Thanks! (Hug)

You can apply the well known trigonometric formula $\displaystyle \sin 2 x = 2\ \sin x\ \cos x$ in a special way writing...

$\displaystyle \sin x = 2\ \sin \frac{x}{2}\ \cos \frac{x}{2} = 4\ \cos \frac{x}{2} \cos \frac{x}{4}\ \sin \frac{x}{4} = ... \(1)$

... and after n interation You obtain...

$\displaystyle \sin x = 2^{n}\ \sin \frac{x}{2^{n}}\ \prod_{k=1}^{n} \cos \frac{x}{2^{k}}\ (2)$

If n tends to infinity the term $2^{n}\ \sin \frac{x}{2^{n}}$ tends to x, so that You obtain...

$\displaystyle \frac{\sin x}{x} = \prod_{k=1}^{\infty} \cos \frac{x}{2^{k}}\ (3)$

Setting in (3) $x=\frac{\pi}{4}$ You obtain Vieta's formula...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

View attachment 2597

If it can be of interest I am currently writing a book focused on the role of cryptanalysis in World War One. As an introduction I wrote a chapter on the state of the cryptanalytic art during the sixteenth century and in this context emerges the figure of Francois Viète, who as well as the brilliant mathematician was also brilliant cryptanalyst. In 1590, Vieta discovered the key to a Spanish cipher, consisting of more than 500 characters, and this meant that all dispatches in that language which fell into the hands of the French could be easily read. When the news came that the secret writings of Spanish cryptographers were regularly deciphered by the French, the king of Spain, Philip II call directly to the Pope saying that Viete supposed to be 'an arch-fiend in the service of the evil one' and had to be brought before a court ecclesiastical to response of witchcraft. The Pope, who well knew that his cryptanalysts were reading for years without difficulty Spanish secret messages did not agree to the request (Tongueout)...

Kind regards

$\chi$ $\sigma$

 

[DreamWeaver]

Thank you both very, very much! That's just what I needed. (Sun)