Algebraic manipulation

[Drez1985]

Main Question or Discussion Point

How does √(2(2r+r√3) simpilfy to r√(2r+√3) ?

Read about the formula in a book about Jamshid al-Kashi, the great Iranian astronomer who used the above formula to determine pi to 16 decimal places.

All help is much appreciated.

 

[Yayness]

Are you sure you wrote it correctly?
The only two cases when $$\sqrt{2(2r+r\sqrt 3)}=r\sqrt{2r+\sqrt 3}$$, are when r=0, or when $$r=\frac{\sqrt{35+16\sqrt 3}-\sqrt 3}{4}$$.

Did you mean $$\sqrt{r(2r^2+\sqrt 3 r)}=r\sqrt{2r+\sqrt 3}$$ ?

 

[Drez1985]

Yes, I'm sure; I wrote off the formula from a book called "The Age of Genius - 1300-1800" by Michael J. Bradley.

Jamshid al-Kashi used the following trignometric formula (as a way to estimate pi):

C_n=√2(2r+C_(n-1))

where C_1=r√3

By first substituting C_1 into the expression for C_2, supposedly, you will end up with:

C_2 = r√(2r+√3)

Was it any help?

P.s. Sorry for the messy notation, but when pasting from the excel equation writer onto this webpage, strange things happens...