Algebraic manipulation

  • Thread starter Drez1985
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  • #1
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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.
 

Answers and Replies

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

Did you mean [tex]\sqrt{r(2r^2+\sqrt 3 r)}=r\sqrt{2r+\sqrt 3}[/tex] ?
 
  • #3
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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...
 

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