4 ms·
Fascinating, I didn't know the babylonian method -- it sounds equivalent to Newton's.
by nicolapede 9y ago
Fascinating, I didn't know the babylonian method -- it sounds equivalent to Newton's.
- mark-r 9y agoI had always thought of this as Newton's Method too, but that's because I was simply misremembering the fullness of Newton's Method [1]. I'm grateful for the education. https://en.wikipedia.org/wiki/Newton%27s_method https://en.wikipedia.org/wiki/Newton%27s_method
- jostylr 9y agoIt is equivalent when applied to the function f(a) = a^(1/2) - x. Also, one needs to divide by 2 since if a is the square root of x, then a+x/a = a + a = 2a. n-th roots are not quite as nice, but still work fairly simply: ((n-1) a + x/a^(n-1) )/n You can see that the fixed point (a = nth root of x) works out as it should.