Square Root Process Similar to Long Division

09182021, 12:46 PM
Post: #4




RE: Square Root Process Similar to Long Division
(09172021 11:00 PM)jeejohn Wrote: Basically, we're keeping track of 2*X on the left, then we add a digit δ to get 2X+δ. Finally we multiple by δ to get 2Xδ+δ^2. We subtract this from the previous quotient to get a new quotient. Above is based on identity: (X+δ)^2 = X^2 + δ*(2X+δ) N = 12345 = (X+δ)^2 X = 100 R = NX^2 = 2345 R = δ*(2X+δ), but we cannot get δ without square root. Newton's method assumed (2X+δ) ≈ 2X, then solve for δ With this assumption, solved δ always overestmate. To compensate, we floored estimated result. R/2/X = 11.725 δ = 11 R = δ*(2X+δ) = 24 X = X+δ = 111 R/2/X ≈ 0.108108108 δ = 0.1081 R = δ*(2X+δ) = 0.00988561 X = X+δ = 111.1081 R/2/X ≈ 4.448645058E5 δ = 4.448646E5 // floor of negative number, same sig. digits as X √12345 ≈ 111.1081 + (4.448646E5) = 111.10805551354 

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Messages In This Thread 
Square Root Process Similar to Long Division  jeejohn  09172021, 11:00 PM
RE: Square Root Process Similar to Long Division  Albert Chan  09172021, 11:58 PM
RE: Square Root Process Similar to Long Division  Albert Chan  09182021, 01:51 AM
RE: Square Root Process Similar to Long Division  Albert Chan  09182021 12:46 PM
RE: Square Root Process Similar to Long Division  Albert Chan  09182021, 01:07 PM
RE: Square Root Process Similar to Long Division  jeejohn  09182021, 08:55 PM

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