／var／logmarcus chiu

Newton's Method - Square Root

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  • π‘₯𝑛+1 = (1/2)Β·(π‘₯𝑛 + (1/π‘₯𝑛))

be equivalent to Newton’s method to find a root of 𝑓(π‘₯) = π‘₯2 βˆ’ π‘Ž. Recall that Newton’s method finds an approximate root of 𝑓(π‘₯) = 0 from a guess π‘₯𝑛 by approximating 𝑓(π‘₯) as its tangent line 𝑓(π‘₯𝑛) + 𝑓’(π‘₯𝑛) (π‘₯ βˆ’ π‘₯𝑛), leading to an improved guess π‘₯𝑛+1Β from the root of the tangent:

  • π‘₯𝑛+1 = π‘₯𝑛 βˆ’ 𝑓(π‘₯𝑛)/𝑓’(π‘₯𝑛),

and for 𝑓(π‘₯) = π‘₯2 βˆ’ π‘Ž this yields the Babylonian formula above