小编Ani*_*der的帖子

我在 Haskell 中有三个函数。所有这些都旨在基于 n 次迭代的初始猜测来执行 \xe2\x88\x9a2 。

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1. \n

   squareRootTwo :: Double -> Integer -> Double\nsquareRootTwo guess n\n| n == 0 = guess\n| otherwise = squareRootTwo ((guess + 2/guess) / 2) (n-1)\n

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2. \n

   squareRootTwoA :: Double -> Integer -> Double\nsquareRootTwoA guess n\n| n == 0 = guess\n| otherwise = squareRootTwoA ((guess + 2/guess) / 2) (n-1) where n=n\n

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3. \n

   squareRootTwoB :: Double -> Integer -> Double\nsquareRootTwoB guess n\n| n == 0 = guess\n| otherwise = …

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