chore: replace comment explaining need for large tolerance

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---

Author: Planeshifter

lib/node_modules/@stdlib/stats/base/dists/bradford/skewness/test/test.js

@@ -80,14 +80,7 @@ tape( 'the function returns the skewness of a bradford distribution', function t
 			delta = abs( y - expected[ i ] );
 
 			/*
-			* NOTE: the tolerance is set high in this case due to the numerically challenging nature of the Bradford distribution skewness formula, which involves:
-			*
-			* 1. Complex expressions with nested logarithmic terms ln(1+c)
-			* 2. Square roots in both numerator and denominator
-			* 3. Products and differences of terms involving c and ln(1+c) that can be sensitive to floating-point precision
-			* 4. The SQRT2 factor amplifying any accumulated numerical errors
-			*
-			* Out of 1000 test cases, only two require tolerance higher than 500*EPS (specifically c=0.4 needs ~20000*EPS).
+			* TODO: the significant divergence from SciPy appears to stem from the computation of the natural log. We should follow up to ensure that our ln implementation is sufficiently accurate.
 			*/
 			tol = 20000.0 * EPS * abs( expected[ i ] );
 			t.ok( delta <= tol, 'within tolerance. x: '+x[i]+'. y: '+y+'. E: '+expected[ i ]+'. Δ: '+delta+'. tol: '+tol+'.' );

lib/node_modules/@stdlib/stats/base/dists/bradford/skewness/test/test.native.js

@@ -89,14 +89,7 @@ tape( 'the function returns the skewness of a Bradford distribution', opts, func
 			delta = abs( y - expected[ i ] );
 
 			/*
-			* NOTE: the tolerance is set high in this case due to the numerically challenging nature of the Bradford distribution skewness formula, which involves:
-			*
-			* 1. Complex expressions with nested logarithmic terms ln(1+c)
-			* 2. Square roots in both numerator and denominator
-			* 3. Products and differences of terms involving c and ln(1+c) that can be sensitive to floating-point precision
-			* 4. The SQRT2 factor amplifying any accumulated numerical errors
-			*
-			* Out of 1000 test cases, only two require tolerance higher than 500*EPS (specifically c=0.4 needs ~20000*EPS).
+			* TODO: the significant divergence from SciPy appears to stem from the computation of the natural log. We should follow up to ensure that our ln implementation is sufficiently accurate.
 			*/
 			tol = 20000.0 * EPS * abs( expected[ i ] );
 			t.ok( delta <= tol, 'within tolerance. c: '+c[i]+'. y: '+y+'. E: '+expected[ i ]+'. Δ: '+delta+'. tol: '+tol+'.' );