The value returned by fnormal is an expression equivalent to e under the assumption that all numeric values with magnitude less than epsilon may be considered to be zero.

In addition, all floats in e which remain nonzero are converted to floats with digits precision.

fnormal preserves numeric type and sign information as much as possible.  Thus, for example, fnormal(1e-20*I, 10) = 0.*I not 0. .  This ensures that branching behavior is generally not affected by fnormal.  Use simplify(expr, zero) to remove 0 real or imaginary parts of complex floating point numbers.

If e is a list, set, range, series, equation, or relation, then fnormal is applied recursively to the components of e.

$\mathrm{fnormal}\left(1.\times {10}^{\mathrm{-10}}\right)$

$0.$

$\mathrm{fnormal}\left(1+1.\times {10}^{\mathrm{-10}}\mathrm{I}\right)$

$1.+0.\mathrm{I}$

$\mathrm{fnormal}\left(\left[{10}^{-10}\,-{10}^{-9}\,{10}^{-8}\right]\right)$

$\left[0.\,\mathrm{-0.}\,\frac{1}{100000000}\right]$

$e≔2x^2+0.00001x-\mathrm{\pi }$

$e≔2x^2+0.00001x-\mathrm{\pi }$

$\mathrm{fnormal}\left(e\right)$

$2x^2+0.00001x-\mathrm{\pi }$

$\mathrm{fnormal}\left(e\,4\right)$

$2x^2-3.141592654$

$f≔\mathrm{evalf}\left(e\right)$

$f≔2.x^2+0.00001x-3.141592654$

$\mathrm{fnormal}\left(f\,4\right)$

$2.x^2-3.142$

$\ln \left(-1.-1.\times {10}^{\mathrm{-15}}\mathrm{I}\right),\ln \left(\mathrm{fnormal}\left(-1.-1.\times {10}^{\mathrm{-15}}\mathrm{I}\,10\right)\right),\ln \left(\mathrm{simplify}\left(\mathrm{fnormal}\left(-1.-1.\times {10}^{\mathrm{-15}}\mathrm{I}\,10\right)\,\mathrm{zero}\right)\right)$

$5.\times {10}^{\mathrm{-31}}-3.141592654\mathrm{I},0.-3.141592654\mathrm{I},0.+3.141592654\mathrm{I}$