User:PeterDixon77/sandbox

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In order to apply Newton's method to find the reciprocal of D, it is necessary to find a function f(X) which has a zero at X=1/D. The obvious such function is f(X)=DX-1, but the Newton–Raphson iteration for this is unhelpful since it cannot be computed without already knowing the reciprocal of D. Moreover multiple iterations for refining reciprocal are not possible since higher order derivatives do not exist for f(X). A function which does work is f(X)=1/X-D, for which the Newton–Raphson iteration gives

X_{i+1} = X_i - {f(X_i)\over f'(X_i)} = X_i - {1/X_i - D\over -1/X_i^2} = X_i + X_i(1-DX_i) = X_i(2-DX_i)

which can be calculated from X_i using only multiplication and subtraction, or using two fused multiply–adds.

From a computation point of view the expressions X_{i+1} = X_i + X_i(1-DX_i) and X_{i+1} = X_i(2-DX_i) are not equivalent. To obtain a result with a precision of n bits while making use of the second expression one must compute the product between X_i and (2-DX_i) with double the required precision (2n bits). In contrast the product between X_i and (1-DX_i) need only be computed with a precision of n bits.

If the error is defined as \epsilon_i = D X_i - 1 \,, then

X_i = {1 \over D} (1 + \epsilon_i) \,
\epsilon_{i+1} = - {\epsilon_i}^2 \,.