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copied!<p>Just to get a feel for it, for your range y = 50, x = 0 gives r = 50 and y = 50, x = +/- 30 gives r ~= 58.3. You want an approximation good for +/- 0.1%, or +/- 0.05 absolute. That's a lot lower accuracy than most library sqrts do.</p> <p>Two approximate approaches - you calculate r based on interpolating from the previous value, or use a few terms of a suitable series. </p> <h2>Interpolating from previous r</h2> <p>r = ( x<sup>2</sup> + y<sup>2</sup> ) <sup>1/2</sup></p> <p>dr/dx = 1/2 . 2x . ( x<sup>2</sup> + y<sup>2</sup> ) <sup>-1/2</sup> = x/r</p> <pre><code> double r = 50; for ( int x = 0; x <= 30; ++x ) { double r_true = Math.sqrt ( 50*50 + x*x ); System.out.printf ( "x: %d r_true: %f r_approx: %f error: %f%%\n", x, r, r_true, 100 * Math.abs ( r_true - r ) / r ); r = r + ( x + 0.5 ) / r; } </code></pre> <p>Gives:</p> <pre><code>x: 0 r_true: 50.000000 r_approx: 50.000000 error: 0.000000% x: 1 r_true: 50.010000 r_approx: 50.009999 error: 0.000002% .... x: 29 r_true: 57.825065 r_approx: 57.801384 error: 0.040953% x: 30 r_true: 58.335225 r_approx: 58.309519 error: 0.044065% </code></pre> <p>which seems to meet the requirement of 0.1% error, so I didn't bother coding the next one, as it would require quite a bit more calculation steps.</p> <h1>Truncated Series</h1> <p>The taylor series for sqrt ( 1 + x ) for x near zero is </p> <p>sqrt ( 1 + x ) = 1 + 1/2 x - 1/8 x<sup>2</sup> ... + ( - 1 / 2 )<sup>n+1</sup> x<sup>n</sup></p> <p>Using r = y sqrt ( 1 + (x/y)<sup>2</sup> ) then you're looking for a term t = ( - 1 / 2 )<sup>n+1</sup> 0.36<sup>n</sup> with magnitude less that a 0.001, log ( 0.002 ) > n log ( 0.18 ) or n > 3.6, so taking terms to x^4 should be Ok.</p>

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