Problem: Investigate following equality:

$\displaystyle \frac{2}{\pi }=\frac{\sqrt{2}}{2} \frac{\sqrt{2+\sqrt{2}}}{2} \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \text{...}$

Thanks.

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- Aug 25th 2012, 10:32 PMMaxJasperInvestigate Equality
Problem: Investigate following equality:

$\displaystyle \frac{2}{\pi }=\frac{\sqrt{2}}{2} \frac{\sqrt{2+\sqrt{2}}}{2} \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \text{...}$

Thanks. - Aug 25th 2012, 11:39 PMVlasevRe: Investigate Equality
What is there to investigate? Do you want to prove the equality? Do you want to formulate a partial product and see if it converges?

- Aug 26th 2012, 09:14 AMMaxJasperRe: Investigate Equality
- Aug 26th 2012, 08:08 PMVlasevRe: Investigate Equality
My thorough investigation revealed that this is Vičte's formula. Now what?

- Aug 26th 2012, 08:24 PMMaxJasperRe: Investigate Equality
Thanks a lot Vlasev, you are truly a good n efficient investigator...good to know the origin of the problem.(Flower)

- Aug 26th 2012, 08:55 PMProve ItRe: Investigate Equality
You can define a circle to be a regular polygon with an infinite number of sides. We can define $\displaystyle \displaystyle \begin{align*} \pi \end{align*}$ as the circumference of a circle of diameter equal to 1 unit. Therefore, we can get an approximation for $\displaystyle \displaystyle \begin{align*} \pi \end{align*}$ by evaluating the perimeter of said regular polygon.

http://upload.wikimedia.org/wikipedi...rolled-720.gif

http://i22.photobucket.com/albums/b3...sson7part2.jpg

http://i22.photobucket.com/albums/b3...dragon/pi1.jpg

http://i22.photobucket.com/albums/b3...sson7part3.jpg

http://i22.photobucket.com/albums/b3...dragon/pi2.jpg

http://i22.photobucket.com/albums/b3...sson7part4.jpg

http://i22.photobucket.com/albums/b3...dragon/pi3.jpg

http://i22.photobucket.com/albums/b3...sson7part5.jpg

Some algebraic manipulation of this will give the result you are trying to prove. - Aug 26th 2012, 09:23 PMMaxJasperRe: Investigate Equality
This explanation is truly enlightening. Thanks a lot ProveIt.

- Aug 26th 2012, 10:42 PMVlasevRe: Investigate Equality
That's indeed a magnificent proof!