Compact form for an infinite multiplication

[cncnewbee]

Hi,
In the middle of the article about Franciscus Vieta, here:
http://en.wikipedia.org/wiki/Franciscus_Vieta

I see an infinite product as an expression for Pi:
2 * 2/2^(1/2) * 2/(2+(2^(1/2))^(1/2) * ...

I was wondering, how this could be written in compact form using math notation please?

 

[SammyS]

Hi,
In the middle of the article about Franciscus Vieta, here:
http://en.wikipedia.org/wiki/Franciscus_Vieta

I see an infinite product as an expression for Pi:
2 * 2/2^(1/2) * 2/(2+(2^(1/2))^(1/2) * ...

I was wondering, how this could be written in compact form using math notation please?

I don't think you can write \displaystyle 2\times\frac{2}{\sqrt{2}}\times\frac{2}{\sqrt{2+ \sqrt{2}}}\times\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\times\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}\times\cdots in compact form.

 

[aznkid66]

Indeed, I don't think any specific notation can help simplify this ... er ... "continued fraction."

I would define the recursive sequence S_n such that:

<br /> S_0 = 1\\<br /> S_1 = \sqrt2\\<br /> S_{k+1} = \sqrt{2+S_k}\text{ where } k&gt;0\\<br />

and use that sequence and capital pi notation to shorten the equation for \pi into the infinite product:

<br /> \pi=\prod \limits_{i=0}^{\infty} \frac{2}{S_i}<br />

But then again, this seems more convoluted than compact.

 

[cncnewbee]

Thank you all, I learned from your answers that the other way of expressing that is by using a recursive expression combined together with Pi (∏) notation and that there is no unique tool in math-notation for a compact form in this case.

 

[Studiot]

I don't know if you want this particular formula for pi or any infinite product will do.

If another is acceptable Google Wallis product.