This is a write up to describe an algorithm described in an ancient Indian manuscript. Its called the Bhakshali manuscript. The manuscript describes some mathematical assertions, methods and algorithms that has been dated to several thousands years ago.

A really cool algorithm described in that manuscript is an approximation for finding the square root of a number. What I liked about this algorithm is that its handy. You could quickly approximate the square root of a real number with just some basic division and addition.

This is how the algorithm works:

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A detailed write up of the algorithm and other dating techniques used can be found here

A really cool algorithm described in that manuscript is an approximation for finding the square root of a number. What I liked about this algorithm is that its handy. You could quickly approximate the square root of a real number with just some basic division and addition.

This is how the algorithm works:

- If 'X' is the number you want to find a square root of, find the nearest whole number 'N' that approximates it. So if X = 23.2 then N = 5.
- Find the difference between X and N*N. Call it D. In this case it works out to -1.8. This should be too tedious to work out either.
- Now comes the magical part, divide this difference (D) by 2*N. So that's -1.8/10. Again, this shouldn't be that difficult to do in your head, -0.18.
- The approximate value of the square root of X is simply N + D/2N = 5 - 0.18 = 4.82

The true value for an approximation of the square root of 23.2 is 4.8166, very close...

A detailed write up of the algorithm and other dating techniques used can be found here

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ReplyDeletethis algorithm comes directly from the taylor expansion of sqrt(1+D), here it is how it works. using the paarameters above.

ReplyDeletefor small D the taylor expansion is: sqrt(1+D)=~1+0.5*D

so if we want to take the square root of X we calculate as:

sqrt(X)=

sqrt(X+N^2-N^2)=

sqrt(N^2+(X-N^2))=

N*sqrt(1+(X-N^2)/(N^2)) =~

N*(1+0.5*(X-N^2)/(N^2))=

N+0.5*(X-N^2)/N=

N+(X-N^2)/(2*N)=

N+D/(2*N)

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