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1 ln(x) e x Approximating functions Elementary Functions

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2 x (i+1) = x (i) (1 +d i 2 –i ) y (i+1) = y (i) – ln(1 +d i 2 –i ) Select d i { 1, 0, 1} s.t. x (m) 1 x (0) =x; y (0) =y x (m) = x (1 +d i 2 –i ) 1 y (m) = y ln(1 +d i 2 –i )= y ln (1 +d i 2 –i ) Let y (0) = 0, x (0) =1; then y (m) = ln(1/x) =lnx Multiplicative Normalization

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3 Range 1/ (1+2 –i ) x 1/ (1 – 2 –i ) 0.21 x 3.45 If the precision of x is k k iterations ln(1±2 –i ) (1 ±2 –i ) for i large x=2 E s with 1 s <2 (IEEE Std: FLP) log 2 x= E+ log 2 e lns= E+1.442685 lns ln x=E ln2+lns = 0.693147 E + lns ln(x)

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4 x (i+1) = x (i) – ln(1 +d i 2 –i ) y (i+1) = y (i) (1 +d i 2 –i ) Select d i { 1, 0, 1} s.t. x (m) 0 x (0) =x; y (0) =y x (m) = x ln(1 +d i 2 –i )= x ln (1 +d i 2 –i ) 0 y (m) = y (1 +d i 2 –i ) ye x Additive Normalization

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5 x (i+1) = x (i) – ln(1 +d i 2 –i ) y (i+1) = y (i) (1 +d i 2 –i ) with x (0) =x; y (0) =1 y (m) e x Domain of convergence 1/ ln(1+2 –i ) x 1/ ln(1 – 2 –i ) 1.25 x 1.56 exex

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6 Taylor series f(x) = Approximating functions

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a) y = 3 x b) y = -3 x c) y = (1/2) x d) y = -(1/2) x.

a) y = 3 x b) y = -3 x c) y = (1/2) x d) y = -(1/2) x.

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