Hi, Le 16/07/2012 19:14, marlon orlando barahona alvarez a écrit : > when we try to calculate the cube root of -27 (using the formula = > POWER (A1, 1/3) -3 Calc should give us but gives us the message # VALUE! > I solved the problem by modifying the formula as follows: > =-POWER (-A1, 1/3) and this modification Calc can give us the correct > answer. > Well I think to include this modification in Cal would take a decision > block, suppose that the entries are "numbers, index". The terms of the > decision would be: > if (num <0 & 1/index% 2 = 1) > The first one test if num is a negative number, the second one trys if > index is an odd number, yet if an integer. If true both conditions > would be used if the proposed amendment and was not used the normal > function. > If this helps please let me know. I hope not look like a fool. If Cal > can do the calculation differently let me know as it does, thanks for > everything. I think it should be better to define a root() function specifically dedicated to compute the root(s). Indeed we should not do confusion between exponentiation and root(s) extraction. x^y is mathematically defined for real numbers and only for x > 0 : x^y = exp(y.Ln(x)). Root extraction is another problem : number of roots, make difference between real and complex root, one argument is real and another is integer.
Best regards. JBF -- Seuls des formats ouverts peuvent assurer la pérennité de vos documents.
_______________________________________________ LibreOffice mailing list LibreOffice@lists.freedesktop.org http://lists.freedesktop.org/mailman/listinfo/libreoffice
