# Algebraic manipulation

## 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.

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

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...